Some Basic concepts of chemistry-Exercises

Q1. Calculate the molar mass of the following:
(i) \(\mathrm{H_2O}\)
(ii) \(\mathrm{CO_2}\)
(iii) \(\mathrm{CH_4}\)

Solution

To calculate molar mass, we add the atomic masses of all atoms present in one molecule. The standard atomic masses used are: \(\mathrm{H = 1\,g\,mol^{-1}},\ \mathrm{C = 12\,g\,mol^{-1}},\ \mathrm{O = 16\,g\,mol^{-1}}\).

(i) For \(\mathrm{H_2O}\):

\[ \begin{aligned} \text{Molar mass of } \mathrm{H_2O} &= 2(\text{H}) + 1(\text{O}) \\ &= 2(1) + 16 \\ &= 2 + 16 \\ &= 18\,\text{g mol}^{-1} \end{aligned} \]

Therefore, the molar mass of water is \(18\,\text{g mol}^{-1}\).

(ii) For \(\mathrm{CO_2}\):

\[ \begin{aligned} \text{Molar mass of } \mathrm{CO_2} &= 1(\text{C}) + 2(\text{O}) \\ &= 12 + 2(16) \\ &= 12 + 32 \\ &= 44\,\text{g mol}^{-1} \end{aligned} \]

Hence, the molar mass of carbon dioxide is \(44\,\text{g mol}^{-1}\).

(iii) For \(\mathrm{CH_4}\):

\[ \begin{aligned} \text{Molar mass of } \mathrm{CH_4} &= 1(\text{C}) + 4(\text{H}) \\ &= 12 + 4(1) \\ &= 12 + 4 \\ &= 16\,\text{g mol}^{-1} \end{aligned} \]

Thus, the molar mass of methane is \(16\,\text{g mol}^{-1}\).

Q2. Calculate the mass percent of different elements present in sodium sulphate \((Na_2SO_4)\).

Solution

To determine the mass percent of each element in sodium sulphate \(\mathrm{Na_2SO_4}\), we first calculate its molar mass using standard atomic masses: \(\mathrm{Na = 23\,g\,mol^{-1}},\ \mathrm{S = 32\,g\,mol^{-1}},\ \mathrm{O = 16\,g\,mol^{-1}}\).

\[ \begin{aligned} \text{Molar mass of } \mathrm{Na_2SO_4} &= 2(\mathrm{Na}) + 1(\mathrm{S}) + 4(\mathrm{O}) \\ &= 2(23) + 32 + 4(16) \\ &= 46 + 32 + 64 \\ &= 142\,\text{g mol}^{-1} \end{aligned} \]

Now, the mass percent of each element is calculated using the relation \(\text{Mass percent} = \dfrac{\text{mass of element in one mole of compound}}{\text{molar mass of compound}} \times 100\).

Sodium:

\[ \begin{aligned} \%\ \mathrm{Na} &= \frac{2 \times 23}{142} \times 100 \\ &= \frac{46}{142} \times 100 \\ &= 32.39\% \end{aligned} \]

Sulphur:

\[ \begin{aligned} \%\ \mathrm{S} &= \frac{32}{142} \times 100 \\ &= 22.54\% \end{aligned} \]

Oxygen:

\[ \begin{aligned} \%\ \mathrm{O} &= \frac{4 \times 16}{142} \times 100 \\ &= \frac{64}{142} \times 100 \\ &= 45.07\% \end{aligned} \]

Therefore, sodium sulphate consists of approximately \(32.39\%\) sodium, \(22.54\%\) sulphur, and \(45.07\%\) oxygen by mass.

Q3. Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1% dioxygen by mass.

Solution

To determine the empirical formula of the oxide of iron, the given mass percentages are converted into relative moles by assuming the total mass of the compound to be \(100\,\text{g}\). Thus, the mass of iron is \(69.9\,\text{g}\) and the mass of dioxygen is \(30.1\,\text{g}\). The atomic masses used are \(\mathrm{Fe = 56\,g\,mol^{-1}}\) and \(\mathrm{O = 16\,g\,mol^{-1}}\).

\[ \begin{aligned} \text{Moles of Fe} &= \frac{69.9}{56} \\ &= 1.248 \\ \text{Moles of O} &= \frac{30.1}{16} \\ &= 1.881 \end{aligned} \]

The mole ratio is obtained by dividing each value by the smaller number, \(1.248\).

\[ \begin{aligned} \text{Fe} &= \frac{1.248}{1.248} = 1.00 \\ \text{O} &= \frac{1.881}{1.248} = 1.51 \end{aligned} \]

Since the ratio of oxygen is approximately \(1.5\), both values are multiplied by \(2\) to obtain the simplest whole number ratio.

\[ \begin{aligned} \text{Fe} &: 1 \times 2 = 2 \\ \text{O} &: 1.5 \times 2 = 3 \end{aligned} \]

Hence, the empirical formula of the given oxide of iron is \(\mathrm{Fe_2O_3}\).

Q4. Calculate the amount of carbon dioxide that could be produced when
(i) 1 mole of carbon is burnt in air.
(ii) 1 mole of carbon is burnt in 16 g of dioxygen.
(iii) 2 moles of carbon are burnt in 16 g of dioxygen.

Solution

The combustion of carbon in oxygen is represented by the balanced chemical equation \(\mathrm{C + O_2 \rightarrow CO_2}\). From this equation, it is clear that \(1\) mole of carbon reacts with \(1\) mole of dioxygen to produce \(1\) mole of carbon dioxide.

(i) When \(1\) mole of carbon is burnt in air, dioxygen is present in excess. Therefore, carbon becomes the limiting reactant.

\[ \begin{aligned} 1\,\text{mol C} &\longrightarrow 1\,\text{mol CO}_2 \end{aligned} \]

Hence, \(1\) mole of carbon dioxide is formed, which corresponds to a mass of

\[ \begin{aligned} 1 \times 44 = 44\,\text{g} \end{aligned} \]

So, \(44\,\text{g}\) of \(\mathrm{CO_2}\) is produced.

(ii) When \(1\) mole of carbon is burnt in \(16\,\text{g}\) of dioxygen, we first convert the mass of dioxygen into moles.

\[ \begin{aligned} \text{Moles of } \mathrm{O_2} &= \frac{16}{32} = 0.5\,\text{mol} \end{aligned} \]

Here, \(0.5\) mole of \(\mathrm{O_2}\) can react with only \(0.5\) mole of carbon. Thus, dioxygen is the limiting reactant.

\[ \begin{aligned} 0.5\,\text{mol O}_2 &\longrightarrow 0.5\,\text{mol CO}_2 \end{aligned} \]

Therefore, the amount of carbon dioxide formed is \(0.5\) mole, which is

\[ \begin{aligned} 0.5 \times 44 = 22\,\text{g} \end{aligned} \]

Hence, \(22\,\text{g}\) of \(\mathrm{CO_2}\) is produced.

(iii) When \(2\) moles of carbon are burnt in \(16\,\text{g}\) of dioxygen, the available dioxygen is again

\[ \begin{aligned} \text{Moles of } \mathrm{O_2} = \frac{16}{32} = 0.5\,\text{mol} \end{aligned} \]

Although \(2\) moles of carbon are present, only \(0.5\) mole of \(\mathrm{O_2}\) is available, making dioxygen the limiting reactant.

\[ \begin{aligned} 0.5\,\text{mol O}_2 &\longrightarrow 0.5\,\text{mol CO}_2 \end{aligned} \]

Thus, even in this case, only \(0.5\) mole of carbon dioxide is formed, corresponding to

\[ \begin{aligned} 0.5 \times 44 = 22\,\text{g} \end{aligned} \]

Therefore, the amount of carbon dioxide produced is \(22\,\text{g}\).

Q5. Calculate the mass of sodium acetate \(\mathrm{(CH_3COONa)}\) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is \(\mathrm{82.0245\ g\ mol^{–1}}\).

Solution

To calculate the mass of sodium acetate required, we first use the definition of molarity, which is given by \(\text{Molarity} = \dfrac{\text{number of moles of solute}}{\text{volume of solution in litres}}\).

The volume of the solution is \(500\,\text{mL} = 0.500\,\text{L}\), and the required molarity is \(0.375\,\text{mol L}^{-1}\).

\[ \begin{aligned} \text{Number of moles of sodium acetate} &= \text{Molarity} \times \text{Volume (in L)} \\ &= 0.375 \times 0.500 \\ &= 0.1875\,\text{mol} \end{aligned} \]

Now, the mass of sodium acetate is obtained using \(\text{Mass} = \text{moles} \times \text{molar mass}\).

\[ \begin{aligned} \text{Mass of } \mathrm{CH_3COONa} &= 0.1875 \times 82.0245 \\ &= 15.38\,\text{g} \ (\text{approximately}) \end{aligned} \]

Therefore, about \(15.38\,\text{g}\) of sodium acetate is required to prepare \(500\,\text{mL}\) of a \(0.375\,\text{M}\) aqueous solution.

Q6. Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 g mL–1 and the mass per cent of nitric acid in it being 69%.

Solution

To find the concentration of nitric acid in moles per litre, we first calculate the mass of the given solution corresponding to \(1\,\text{L}\). Since the density of the solution is \(1.41\,\text{g mL}^{-1}\), the mass of \(1000\,\text{mL}\) of solution is:

\[ \begin{aligned} \text{Mass of solution} &= 1.41 \times 1000 \\ &= 1410\,\text{g} \end{aligned} \]

The mass percent of nitric acid is \(69\%\), which means that \(69\,\text{g}\) of \(\mathrm{HNO_3}\) is present in every \(100\,\text{g}\) of solution. Therefore, the mass of nitric acid in \(1410\,\text{g}\) of solution is:

\[ \begin{aligned} \text{Mass of } \mathrm{HNO_3} &= \frac{69}{100} \times 1410 \\ &= 972.9\,\text{g} \end{aligned} \]

Now, using the molar mass of nitric acid, \(\mathrm{HNO_3 = 63\,g\,mol^{-1}}\), the number of moles present in this mass is:

\[ \begin{aligned} \text{Moles of } \mathrm{HNO_3} &= \frac{972.9}{63} \\ &= 15.44\,\text{mol} \end{aligned} \]

Since these \(15.44\) moles are present in \(1\,\text{L}\) of solution, the concentration of nitric acid is:

\[ \begin{aligned} \text{Molarity} &= 15.44\,\text{mol L}^{-1} \end{aligned} \]

Hence, the concentration of nitric acid in the given sample is approximately \(15.44\,\text{M}\).

Q7. How much copper can be obtained from 100 g of copper sulphate \(\mathrm{(CuSO_4)}\)?

Solution

To find how much copper can be obtained from \(100\,\text{g}\) of copper sulphate \(\mathrm{(CuSO_4)}\), we first determine the fraction of copper present in one mole of the compound. The atomic masses used are \(\mathrm{Cu = 63.5\,g\,mol^{-1}},\ \mathrm{S = 32\,g\,mol^{-1}}\) and \(\mathrm{O = 16\,g\,mol^{-1}}\).

\[ \begin{aligned} \text{Molar mass of } \mathrm{CuSO_4} &= 63.5 + 32 + 4(16) \\ &= 63.5 + 32 + 64 \\ &= 159.5\,\text{g mol}^{-1} \end{aligned} \]

Thus, \(159.5\,\text{g}\) of \(\mathrm{CuSO_4}\) contains \(63.5\,\text{g}\) of copper. The mass of copper present in \(100\,\text{g}\) of copper sulphate is calculated as:

\[ \begin{aligned} \text{Mass of Cu} &= \frac{63.5}{159.5} \times 100 \\ &= 39.8\,\text{g} \ (\text{approximately}) \end{aligned} \]

Therefore, about \(39.8\,\text{g}\) of copper can be obtained from \(100\,\text{g}\) of copper sulphate.

Q8. Determine the molecular formula of an oxide of iron, in which the mass per cent of iron and oxygen are 69.9 and 30.1, respectively.

Solution

To determine the molecular formula of the oxide of iron, we first convert the given mass percentages into moles by assuming the total mass of the compound to be \(100\,\text{g}\). Hence, iron contributes \(69.9\,\text{g}\) and oxygen contributes \(30.1\,\text{g}\). The atomic masses used are \(\mathrm{Fe = 56\,g\,mol^{-1}}\) and \(\mathrm{O = 16\,g\,mol^{-1}}\).

\[ \begin{aligned} \text{Moles of Fe} &= \frac{69.9}{56} = 1.248 \\ \text{Moles of O} &= \frac{30.1}{16} = 1.881 \end{aligned} \]

Next, the simplest mole ratio is obtained by dividing each value by the smaller number, \(1.248\).

\[ \begin{aligned} \text{Fe} &= \frac{1.248}{1.248} = 1.00 \\ \text{O} &= \frac{1.881}{1.248} = 1.51 \end{aligned} \]

Since the ratio of oxygen is close to \(1.5\), both values are multiplied by \(2\) to convert them into whole numbers.

\[ \begin{aligned} \text{Fe} &: 1 \times 2 = 2 \\ \text{O} &: 1.51 \times 2 \approx 3 \end{aligned} \]

Thus, the empirical formula of the oxide is \(\mathrm{Fe_2O_3}\). As no additional molar mass data is provided, the empirical formula itself represents the molecular formula.

Therefore, the molecular formula of the given oxide of iron is \(\mathrm{Fe_2O_3}\).

Q9. Calculate the atomic mass (average) of chlorine using the following data:

Solution

To calculate the average atomic mass of chlorine, we use the weighted mean formula, where each isotopic mass is multiplied by its fractional natural abundance and then summed.

First, the given percentage abundances are converted into fractions: \(\mathrm{^{35}Cl} = 75.77\% = 0.7577\) and \(\mathrm{^{37}Cl} = 24.23\% = 0.2423\).

The average atomic mass is therefore calculated as:

\[ \begin{aligned} \text{Average atomic mass of Cl} &= (0.7577 \times 34.9689) + (0.2423 \times 36.9659) \\ &= 26.4937 + 8.9565 \\ &= 35.4502 \end{aligned} \]

Thus, the average atomic mass of chlorine is approximately \(35.45\,\text{u}\).

Q10. In three moles of ethane \(\mathrm{(C_2H_6)}\), calculate the following:
(i) Number of moles of carbon atoms.
(ii) Number of moles of hydrogen atoms.
(iii) Number of molecules of ethane.

Solution

Ethane has the molecular formula \(\mathrm{C_2H_6}\), which shows that each molecule contains \(2\) carbon atoms and \(6\) hydrogen atoms. The given amount is \(3\) moles of ethane.

(i) Number of moles of carbon atoms:

\[ \begin{aligned} 1\,\text{mol of } \mathrm{C_2H_6} &\text{ contains } 2\,\text{mol of C atoms} \\ 3\,\text{mol of } \mathrm{C_2H_6} &\text{ contains } 3 \times 2 = 6\,\text{mol of C atoms} \end{aligned} \]

Therefore, \(6\,\text{moles}\) of carbon atoms are present.

(ii) Number of moles of hydrogen atoms:

\[ \begin{aligned} 1\,\text{mol of } \mathrm{C_2H_6} &\text{ contains } 6\,\text{mol of H atoms} \\ 3\,\text{mol of } \mathrm{C_2H_6} &\text{ contains } 3 \times 6 = 18\,\text{mol of H atoms} \end{aligned} \]

Hence, \(18\,\text{moles}\) of hydrogen atoms are present.

(iii) Number of molecules of ethane:

Using Avogadro’s number, \(1\,\text{mol} = 6.022 \times 10^{23}\) molecules.

\[ \begin{aligned} \text{Number of molecules} &= 3 \times 6.022 \times 10^{23} \\ &= 1.8066 \times 10^{24} \end{aligned} \]

Thus, \(3\) moles of ethane contain approximately \(1.81 \times 10^{24}\) molecules.

Q11. What is the concentration of sugar \(\mathrm{(C_{12} H_{22} O_{11} )}\) in \(\mathrm{mol\, L^(–1)}\) if its 20 g are dissolved in enough water to make a final volume up to 2L?

Solution

To find the concentration of sugar in \(\mathrm{mol\,L^{-1}}\), we first calculate the number of moles of sugar present. The molecular formula of sugar is \(\mathrm{C_{12}H_{22}O_{11}}\). Using atomic masses \(\mathrm{C = 12}\), \(\mathrm{H = 1}\), and \(\mathrm{O = 16}\), its molar mass is obtained as follows.

\[ \begin{aligned} \text{Molar mass of } \mathrm{C_{12}H_{22}O_{11}} &= 12(12) + 22(1) + 11(16) \\ &= 144 + 22 + 176 \\ &= 342\,\text{g mol}^{-1} \end{aligned} \]

Now, the number of moles of sugar present in \(20\,\text{g}\) is:

\[ \begin{aligned} \text{Moles of sugar} &= \frac{20}{342} \\ &= 0.0585\,\text{mol (approximately)} \end{aligned} \]

The final volume of the solution is \(2\,\text{L}\). Molarity is defined as moles of solute per litre of solution.

\[ \begin{aligned} \text{Concentration (M)} &= \frac{0.0585}{2} \\ &= 0.0293\,\text{mol L}^{-1} \end{aligned} \]

Therefore, the concentration of sugar in the solution is approximately \(2.93 \times 10^{-2}\,\text{mol L}^{-1}\).

Q12. If the density of methanol is 0.793 kg L–1, what is its volume needed for making 2.5 L of its 0.25 M solution?

Solution

To determine the volume of methanol required, we first calculate the number of moles needed to prepare \(2.5\,\text{L}\) of a \(0.25\,\text{M}\) solution. Using the relation \(\text{moles} = \text{molarity} \times \text{volume}\),

\[ \begin{aligned} \text{Moles of methanol} &= 0.25 \times 2.5 \\ &= 0.625\,\text{mol} \end{aligned} \]

The molar mass of methanol \(\mathrm{(CH_3OH)}\) is calculated using \(\mathrm{C = 12}\), \(\mathrm{H = 1}\), and \(\mathrm{O = 16}\).

\[ \begin{aligned} \text{Molar mass of } \mathrm{CH_3OH} &= 12 + 4(1) + 16 \\ &= 32\,\text{g mol}^{-1} \end{aligned} \]

Now, the mass of methanol corresponding to \(0.625\,\text{mol}\) is:

\[ \begin{aligned} \text{Mass of methanol} &= 0.625 \times 32 \\ &= 20\,\text{g} \end{aligned} \]

The given density of methanol is \(0.793\,\text{kg L}^{-1} = 793\,\text{g L}^{-1}\). Using the relation \(\text{Volume} = \dfrac{\text{mass}}{\text{density}}\),

\[ \begin{aligned} \text{Volume of methanol} &= \frac{20}{793} \\ &= 0.0252\,\text{L} \end{aligned} \]

Converting this into millilitres,

\[ \begin{aligned} 0.0252\,\text{L} = 25.2\,\text{mL} \end{aligned} \]

Therefore, approximately \(25.2\,\text{mL}\) of methanol is required to prepare \(2.5\,\text{L}\) of a \(0.25\,\text{M}\) solution.

Q13. Pressure is determined as force per unit area of the surface. The SI unit of pressure, pascal is as shown below:
\(\mathrm{1\,Pa = 1\,N\, m^{–2}}\)
If mass of air at sea level is \(\mathrm{10^{34}\, g\, cm^{–2}}\), calculate the pressure in pascal

Solution

Pressure is defined as force per unit area. The given data states that the mass of air at sea level is \(10^{3}\,\text{g cm}^{-2}\). To calculate pressure in pascal, this mass per unit area must first be converted into SI units and then multiplied by acceleration due to gravity.

First, the mass per unit area is converted into \(\text{kg m}^{-2}\).

\[ \begin{aligned} 10^{3}\,\text{g cm}^{-2} &= 10^{3} \times 10^{-3}\,\text{kg} \times (10^{4}\,\text{cm}^2\,\text{per m}^2) \\ &= 10^{4}\,\text{kg m}^{-2} \end{aligned} \]

Pressure is given by the product of mass per unit area and acceleration due to gravity. Taking \(g = 9.8\,\text{m s}^{-2}\),

\[ \begin{aligned} \text{Pressure} &= (10^{4}\,\text{kg m}^{-2}) \times (9.8\,\text{m s}^{-2}) \\ &= 9.8 \times 10^{4}\,\text{N m}^{-2} \end{aligned} \]

Since \(1\,\text{Pa} = 1\,\text{N m}^{-2}\), the pressure at sea level is \[ 9.8 \times 10^{4}\,\text{Pa}. \]

Thus, the pressure exerted by air at sea level is approximately \(9.8 \times 10^{4}\,\text{Pa}\).

Q14. What is the SI unit of mass? How is it defined?

Solution

The SI unit of mass is the kilogram, symbol \(\mathrm{kg}\). It is the base unit of mass in the International System of Units (SI) and is used universally for scientific, industrial, and commercial measurements.

The kilogram is defined by fixing the numerical value of the Planck constant \(h\) to be exactly \[ h = 6.62607015 \times 10^{-34}\,\text{J s}, \] where the joule \((\text{J})\) is expressed in terms of the SI base units \(\mathrm{kg\, m^2\, s^{-1}}\).

With this definition, the kilogram is no longer based on a physical object but is realized through precise measurements involving fundamental constants of nature, ensuring long-term stability and universal reproducibility.

Q15. Match the following prefixes with their multiples: \[ \begin{array}{c|c} \begin{aligned} &\text{Prefixes}&\text{Multiples}\\ (i)\quad&\text{micro}&10^6\\ (ii)\quad&\text{deca}&10^9\\ (iii)\quad&\text{mega}&10^{-6}\\ (iv)\quad&\text{giga}&10^{-15}\\ (v)\quad&\text{femto}&10 \end{aligned} \end{array} \]

Solution

To match each prefix with its correct multiple, we recall the standard SI meanings of the given prefixes. Micro represents one millionth, deca represents ten times, mega represents one million, giga represents one billion, and femto represents one quadrillionth.

Using these definitions, the correct matches are obtained as follows.

\[ \begin{aligned} \text{micro} &\rightarrow 10^{-6} \\ \text{deca} &\rightarrow 10 \\ \text{mega} &\rightarrow 10^{6} \\ \text{giga} &\rightarrow 10^{9} \\ \text{femto} &\rightarrow 10^{-15} \end{aligned} \]

Hence, the correct pairing is: (i) micro – \(10^{-6}\), (ii) deca – \(10\), (iii) mega – \(10^{6}\), (iv) giga – \(10^{9}\), and (v) femto – \(10^{-15}\).

16. What do you mean by significant figures?

Solution

Significant figures are the meaningful digits in a measured or calculated quantity that indicate the precision of that value. They include all certain digits obtained directly from measurement along with the first uncertain (estimated) digit.

In any numerical result, significant figures convey how reliable the measurement is. For example, in the value \(2.35\), all three digits are significant because each contributes to the accuracy of the measurement. On the other hand, in \(0.0042\), only \(4\) and \(2\) are significant, since the leading zeros merely locate the decimal point and do not carry information about precision.

Thus, significant figures help express both the magnitude of a quantity and the degree of certainty associated with its measurement.

Q17. A sample of drinking water was found to be severely contaminated with chloroform, \(\mathrm{CHCl_3}\) , supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass).
(i) Express this in per cent by mass.
(ii) Determine the molality of chloroform in the water sample

Solution

The contamination level of chloroform in drinking water is given as \(15\) ppm by mass. This means that \(15\) parts of chloroform are present in \(10^{6}\) parts of water by mass.

(i) To express this value in percent by mass, we use the relation \(\text{percent} = \dfrac{\text{ppm}}{10^{4}}\).

\[ \begin{aligned} \text{Mass percent of } \mathrm{CHCl_3} &= \frac{15}{10^{4}} \\ &= 1.5 \times 10^{-3}\% \end{aligned} \]

Hence, the concentration of chloroform is \(1.5 \times 10^{-3}\%\) by mass.

(ii) To calculate molality, we again assume \(1\,\text{kg}\) of water (solvent). At \(15\) ppm, this corresponds to \(15\,\text{mg}\) of chloroform dissolved in \(1\,\text{kg}\) of water.

\[ \begin{aligned} 15\,\text{mg} &= 0.015\,\text{g} \end{aligned} \]

The molar mass of chloroform \(\mathrm{(CHCl_3)}\) is:

\[ \begin{aligned} 12 + 1 + 3(35.5) &= 119.5\,\text{g mol}^{-1} \end{aligned} \]

Now, the number of moles of chloroform present is:

\[ \begin{aligned} \text{Moles of } \mathrm{CHCl_3} &= \frac{0.015}{119.5} \\ &= 1.26 \times 10^{-4}\,\text{mol} \end{aligned} \]

Since this amount is present in \(1\,\text{kg}\) of water, the molality is:

\[ \begin{aligned} \text{Molality} &= 1.26 \times 10^{-4}\,\text{mol kg}^{-1} \end{aligned} \]

Therefore, the contamination corresponds to \(1.5 \times 10^{-3}\%\) by mass and a molality of approximately \(1.26 \times 10^{-4}\,\text{m}\).

Q18. Express the following in the scientific notation:
(i) 0.0048
(ii) 234,000
(iii) 8008
(iv) 500.0
(v) 6.0012 

Solution

Scientific notation expresses a number in the form \(a \times 10^{n}\), where \(1 \le a < 10\) and \(n\) is an integer. Each given value is rewritten by shifting the decimal point appropriately while keeping the significant figures unchanged.

For \(0.0048\), the decimal point is moved three places to the right.

\[ \begin{aligned} 0.0048 &= 4.8 \times 10^{-3} \end{aligned} \]

For \(234{,}000\), the decimal point is moved five places to the left.

\[ \begin{aligned} 234000 &= 2.34 \times 10^{5} \end{aligned} \]

For \(8008\), the decimal point is shifted three places to the left.

\[ \begin{aligned} 8008 &= 8.008 \times 10^{3} \end{aligned} \]

For \(500.0\), keeping the given significant figures, the decimal point is moved two places to the left.

\[ \begin{aligned} 500.0 &= 5.000 \times 10^{2} \end{aligned} \]

For \(6.0012\), the number already lies between \(1\) and \(10\), so no shift is needed.

\[ \begin{aligned} 6.0012 &= 6.0012 \times 10^{0} \end{aligned} \]

Thus, the scientific notation forms are \(4.8 \times 10^{-3}\), \(2.34 \times 10^{5}\), \(8.008 \times 10^{3}\), \(5.000 \times 10^{2}\), and \(6.0012 \times 10^{0}\), respectively.

Q19. How many significant figures are present in the following?
(i) 0.0025
(ii) 208
(iii) 5005
(iv) 126,000
(v) 500.0
(vi) 2.0034

Solution

The number of significant figures in a measured quantity depends on which digits carry meaningful information about precision. Leading zeros are not significant, zeros between non-zero digits are significant, and trailing zeros are significant only when a decimal point is explicitly shown.

For \(0.0025\), the leading zeros only locate the decimal point and do not count. Only the digits \(2\) and \(5\) are significant.

\[ \begin{aligned} 0.0025 &\rightarrow 2 \text{ significant figures} \end{aligned} \]

In the number \(208\), the zero lies between two non-zero digits and therefore is significant along with the other digits.

\[ \begin{aligned} 208 &\rightarrow 3 \text{ significant figures} \end{aligned} \]

In \(5005\), both zeros are sandwiched between non-zero digits, so all digits contribute to the precision.

\[ \begin{aligned} 5005 &\rightarrow 4 \text{ significant figures} \end{aligned} \]

For \(126{,}000\), no decimal point is given, so the trailing zeros are not taken as significant. Only the digits \(1\), \(2\), and \(6\) are counted.

\[ \begin{aligned} 126000 &\rightarrow 3 \text{ significant figures} \end{aligned} \]

In \(500.0\), the presence of the decimal point indicates that the trailing zeros are significant and reflect the measured precision.

\[ \begin{aligned} 500.0 &\rightarrow 4 \text{ significant figures} \end{aligned} \]

For \(2.0034\), all digits including the zeros between non-zero digits are significant.

\[ \begin{aligned} 2.0034 &\rightarrow 5 \text{ significant figures} \end{aligned} \]

Q20. Round up the following upto three significant figures:
(i) 34.216
(ii) 10.4107
(iii) 0.04597
(iv) 2808

Solution

Rounding to three significant figures means retaining the first three meaningful digits and then adjusting the last retained digit based on the value of the next digit to its right.

For the number \(34.216\), the first three significant figures are \(3\), \(4\), and \(2\). The next digit is \(1\), which is less than \(5\), so no rounding up is required.

\[ \begin{aligned} 34.216 &\approx 34.2 \end{aligned} \]

In the case of \(10.4107\), the first three significant figures are \(1\), \(0\), and \(4\). The next digit is \(1\), which does not change the third significant figure.

\[ \begin{aligned} 10.4107 &\approx 10.4 \end{aligned} \]

For \(0.04597\), leading zeros are not significant. The first three significant figures are \(4\), \(5\), and \(9\). The next digit is \(7\), which is greater than \(5\), so the third significant figure is increased by one.

\[ \begin{aligned} 0.04597 &\approx 0.0460 \end{aligned} \]

In the number \(2808\), all digits are significant. The first three significant figures are \(2\), \(8\), and \(0\). The next digit is \(8\), so the third significant figure is rounded up.

\[ \begin{aligned} 2808 &\approx 2.81 \times 10^{3} \end{aligned} \]

Thus, the rounded values up to three significant figures are \(34.2\), \(10.4\), \(0.0460\), and \(2.81 \times 10^{3}\), respectively.

Q21. The following data are obtained when dinitrogen and dioxygen react together to form different compounds: \[ \begin{array}{c|c} \begin{aligned} &\text{Mass of dinitrogen}&\text{Mass of dioxygen}\\ (i)\quad&\text{14 g}&\text{16 g}\\ (ii)\quad&\text{14 g}&\text{32 g}\\ (iii)\quad&\text{28 g}&\text{32 g}\\ (iv)\quad&\text{28 g}&\text{80 g} \end{aligned} \end{array} \] (a) Which law of chemical combination is obeyed by the above experimental data? Give its statement.
(b) Fill in the blanks in the following conversions:
(i) 1 km = ...................... mm = ...................... pm
(ii) 1 mg = ...................... kg = ...................... ng
(iii) 1 mL = ...................... L = ...................... dm3

Solution

From the given data, different compounds are formed by the combination of dinitrogen and dioxygen using fixed masses of nitrogen while the masses of oxygen vary in simple numerical ratios. This observation directly points to a fundamental law of chemical combination.

The experimental data obey the law of multiple proportions. According to this law, when two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element are in the ratio of small whole numbers.

In the given cases, when \(14\,\text{g}\) of nitrogen combines with oxygen, the masses of oxygen are \(16\,\text{g}\) and \(32\,\text{g}\), which are in the simple ratio \(1:2\). Similarly, when \(28\,\text{g}\) of nitrogen reacts, the corresponding masses of oxygen are \(32\,\text{g}\) and \(80\,\text{g}\), giving the ratio \(2:5\). These simple whole-number ratios confirm the law of multiple proportions.

Now, the required unit conversions are completed using standard SI relations.

For length conversion:

\[ \begin{aligned} 1\,\text{km} &= 10^{6}\,\text{mm} \\ &= 10^{15}\,\text{pm} \end{aligned} \]

For mass conversion:

\[ \begin{aligned} 1\,\text{mg} &= 10^{-6}\,\text{kg} \\ &= 10^{6}\,\text{ng} \end{aligned} \]

For volume conversion:

\[ \begin{aligned} 1\,\text{mL} &= 10^{-3}\,\text{L} \\ &= 10^{-3}\,\text{dm}^{3} \end{aligned} \]

Thus, the given data illustrate the law of multiple proportions, and the unit conversions are completed using standard SI relationships.

Q22. If the speed of light is \(\mathrm{3.0 \times 10^8\ m\ s^{–1}}\), calculate the distance covered by light in 2.00 ns.

Solution

The distance covered by light can be calculated using the relation \(\text{distance} = \text{speed} \times \text{time}\). The given speed of light is \(3.0 \times 10^{8}\,\text{m s}^{-1}\).

First, the given time is converted into seconds. Since \(1\,\text{ns} = 10^{-9}\,\text{s}\),

\[ \begin{aligned} 2.00\,\text{ns} &= 2.00 \times 10^{-9}\,\text{s} \end{aligned} \]

Now, substituting the values into the formula,

\[ \begin{aligned} \text{Distance} &= (3.0 \times 10^{8}) \times (2.00 \times 10^{-9}) \\ &= 6.00 \times 10^{-1}\,\text{m} \end{aligned} \]

Thus, the distance covered by light in \(2.00\,\text{ns}\) is \(0.600\,\text{m}\).

Q23. In a reaction \[\mathrm{A + B_2 \rightarrow AB_2}\] Identify the limiting reagent, if any, in the following reaction mixtures.
(i) 300 atoms of A + 200 molecules of B
(ii) 2 mol A + 3 mol B
(iii) 100 atoms of A + 100 molecules of B
(iv) 5 mol A + 2.5 mol B
(v) 2.5 mol A + 5 mol B

Solution

The balanced reaction is \(\mathrm{A + B_2 \rightarrow AB_2}\). This equation shows that one atom (or mole) of \(\mathrm{A}\) reacts with one molecule (or mole) of \(\mathrm{B_2}\). Therefore, the reactants must be present in a \(1:1\) ratio for complete reaction. The limiting reagent is the one present in the lesser stoichiometric amount.

For case (i), \(300\) atoms of \(\mathrm{A}\) are mixed with \(200\) molecules of \(\mathrm{B_2}\). Since the required ratio is \(1:1\), only \(200\) atoms of \(\mathrm{A}\) can react with \(200\) molecules of \(\mathrm{B_2}\).

\[ \begin{aligned} \text{Limiting reagent} &= \mathrm{B_2} \end{aligned} \]

For case (ii), \(2\,\text{mol}\) of \(\mathrm{A}\) react with \(3\,\text{mol}\) of \(\mathrm{B_2}\). According to the equation, \(2\,\text{mol}\) of \(\mathrm{A}\) require only \(2\,\text{mol}\) of \(\mathrm{B_2}\).

\[ \begin{aligned} \text{Limiting reagent} &= \mathrm{A} \end{aligned} \]

For case (iii), \(100\) atoms of \(\mathrm{A}\) are present with \(100\) molecules of \(\mathrm{B_2}\). The quantities are exactly in the required \(1:1\) ratio.

\[ \begin{aligned} \text{Limiting reagent} &= \text{None (both are completely consumed)} \end{aligned} \]

For case (iv), \(5\,\text{mol}\) of \(\mathrm{A}\) are mixed with \(2.5\,\text{mol}\) of \(\mathrm{B_2}\). Only \(2.5\,\text{mol}\) of \(\mathrm{A}\) can react with the available \(\mathrm{B_2}\).

\[ \begin{aligned} \text{Limiting reagent} &= \mathrm{B_2} \end{aligned} \]

For case (v), \(2.5\,\text{mol}\) of \(\mathrm{A}\) are mixed with \(5\,\text{mol}\) of \(\mathrm{B_2}\). The given amount of \(\mathrm{A}\) requires only \(2.5\,\text{mol}\) of \(\mathrm{B_2}\).

\[ \begin{aligned} \text{Limiting reagent} &= \mathrm{A} \end{aligned} \]

Thus, the limiting reagents are \(\mathrm{B_2}\), \(\mathrm{A}\), none, \(\mathrm{B_2}\), and \(\mathrm{A}\) for cases (i) to (v), respectively.

Q24. Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation: \[N_2 (g) + H_2 (g) \rightarrow 2NH_3 (g)\] (i) Calculate the mass of ammonia produced if 2.00 × 103 g dinitrogen reacts with 1.00 × 103 g of dihydrogen.
(ii) Will any of the two reactants remain unreacted?
(iii) If yes, which one and what would be its mass?

Solution

The correct balanced chemical equation for the formation of ammonia is \(\mathrm{N_2 + 3H_2 \rightarrow 2NH_3}\). This shows that \(1\) mole of dinitrogen reacts with \(3\) moles of dihydrogen to produce \(2\) moles of ammonia.

First, the given masses are converted into moles. The molar masses used are \(\mathrm{N_2 = 28\,g\,mol^{-1}}\), \(\mathrm{H_2 = 2\,g\,mol^{-1}}\), and \(\mathrm{NH_3 = 17\,g\,mol^{-1}}\).

\[ \begin{aligned} \text{Moles of } \mathrm{N_2} &= \frac{2.00 \times 10^{3}}{28} = 71.43\,\text{mol} \\ \text{Moles of } \mathrm{H_2} &= \frac{1.00 \times 10^{3}}{2} = 500\,\text{mol} \end{aligned} \]

According to the balanced equation, \(71.43\) moles of \(\mathrm{N_2}\) would require

\[ \begin{aligned} \text{Required moles of } \mathrm{H_2} &= 3 \times 71.43 = 214.29\,\text{mol} \end{aligned} \]

Since \(500\,\text{mol}\) of \(\mathrm{H_2}\) are available but only \(214.29\,\text{mol}\) are needed, dihydrogen is in excess and dinitrogen is the limiting reagent.

Now, from the stoichiometry, \(1\) mole of \(\mathrm{N_2}\) produces \(2\) moles of \(\mathrm{NH_3}\). Therefore, the moles of ammonia formed are:

\[ \begin{aligned} \text{Moles of } \mathrm{NH_3} &= 2 \times 71.43 = 142.86\,\text{mol} \end{aligned} \]

The corresponding mass of ammonia is:

\[ \begin{aligned} \text{Mass of } \mathrm{NH_3} &= 142.86 \times 17 \\ &= 2.43 \times 10^{3}\,\text{g (approximately)} \end{aligned} \]

Thus, about \(2.43 \times 10^{3}\,\text{g}\) of ammonia is produced.

Since dinitrogen is the limiting reagent, dihydrogen remains unreacted. The amount of dihydrogen actually consumed is:

\[ \begin{aligned} \text{Consumed } \mathrm{H_2} &= 214.29\,\text{mol} \end{aligned} \]

The remaining moles of dihydrogen are therefore:

\[ \begin{aligned} \text{Unreacted } \mathrm{H_2} &= 500 - 214.29 = 285.71\,\text{mol} \end{aligned} \]

Converting this into mass,

\[ \begin{aligned} \text{Mass of unreacted } \mathrm{H_2} &= 285.71 \times 2 \\ &= 5.71 \times 10^{2}\,\text{g} \end{aligned} \]

Therefore, dihydrogen remains in excess, and approximately \(5.71 \times 10^{2}\,\text{g}\) of \(\mathrm{H_2}\) is left unreacted.

Q25. How are 0.50 mol \(\mathrm{Na_2CO_3}\) and 0.50 M \(\mathrm{Na_2CO_3}\) different?

Solution

The expressions \(0.50\,\text{mol } \mathrm{Na_2CO_3}\) and \(0.50\,\text{M } \mathrm{Na_2CO_3}\) represent two different ways of specifying the amount of sodium carbonate.

When we say \(0.50\,\text{mol } \mathrm{Na_2CO_3}\), it refers only to the quantity of substance. It means exactly half a mole of sodium carbonate, regardless of the volume in which it is present. In terms of mass, this corresponds to

\[ \begin{aligned} \text{Mass} &= 0.50 \times 106 \\ &= 53\,\text{g} \end{aligned} \]

where \(106\,\text{g mol}^{-1}\) is the molar mass of \(\mathrm{Na_2CO_3}\). Thus, \(0.50\,\text{mol}\) simply specifies \(53\,\text{g}\) of the compound and does not involve any information about solution volume.

On the other hand, \(0.50\,\text{M } \mathrm{Na_2CO_3}\) describes the concentration of a solution. Molarity is defined as moles of solute per litre of solution, so

\[ \begin{aligned} 0.50\,\text{M} &= 0.50\,\text{mol L}^{-1} \end{aligned} \]

This means that \(0.50\) mole (or \(53\,\text{g}\)) of \(\mathrm{Na_2CO_3}\) is dissolved in enough water to make the total volume of the solution exactly \(1\,\text{L}\). Here, both the amount of solute and the volume of solution are specified.

Therefore, \(0.50\,\text{mol } \mathrm{Na_2CO_3}\) indicates only the quantity of substance, whereas \(0.50\,\text{M } \mathrm{Na_2CO_3}\) indicates the concentration of sodium carbonate in a solution.

Q26. If 10 volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of water vapour would be produced?

Solution

The reaction between dihydrogen and dioxygen to form water vapour is represented by the balanced chemical equation \(\mathrm{2H_2(g) + O_2(g) \rightarrow 2H_2O(g)}\).

According to Gay–Lussac’s law of combining volumes, when gases react at the same temperature and pressure, the volumes of the reacting gases and the gaseous products are in simple whole-number ratios given by the balanced equation.

From the equation, \(2\) volumes of dihydrogen react with \(1\) volume of dioxygen to produce \(2\) volumes of water vapour.

In the given case, \(10\) volumes of dihydrogen react with \(5\) volumes of dioxygen. This ratio simplifies to

\[ \begin{aligned} 10 : 5 &= 2 : 1 \end{aligned} \]

which exactly matches the stoichiometric ratio required by the balanced equation. Hence, both gases react completely without any excess.

Since \(2\) volumes of dihydrogen produce \(2\) volumes of water vapour, \(10\) volumes of dihydrogen will produce

\[ \begin{aligned} \text{Volume of } \mathrm{H_2O(g)} &= 10\,\text{volumes} \end{aligned} \]

Therefore, \(10\) volumes of water vapour would be produced.

Q27. Convert the following into basic units:
(i) 28.7 pm
(ii) 15.15 pm
(iii) 25365 mg

Solution

To convert the given quantities into basic SI units, we use the relations \(1\,\text{pm} = 10^{-12}\,\text{m}\) and \(1\,\text{mg} = 10^{-6}\,\text{kg}\).

For \(28.7\,\text{pm}\),

\[ \begin{aligned} 28.7\,\text{pm} &= 28.7 \times 10^{-12}\,\text{m} \\ &= 2.87 \times 10^{-11}\,\text{m} \end{aligned} \]

For \(15.15\,\text{pm}\),

\[ \begin{aligned} 15.15\,\text{pm} &= 15.15 \times 10^{-12}\,\text{m} \\ &= 1.515 \times 10^{-11}\,\text{m} \end{aligned} \]

For \(25365\,\text{mg}\),

\[ \begin{aligned} 25365\,\text{mg} &= 25365 \times 10^{-6}\,\text{kg} \\ &= 2.5365 \times 10^{-2}\,\text{kg} \end{aligned} \]

Hence, the values in basic SI units are \(2.87 \times 10^{-11}\,\text{m}\), \(1.515 \times 10^{-11}\,\text{m}\), and \(2.5365 \times 10^{-2}\,\text{kg}\), respectively.

Q28. Which one of the following will have the largest number of atoms?
(i) 1 g Au (s)
(ii) 1 g Na (s)
(iii) 1 g Li (s)
(iv) 1 g of \(\mathrm{Cl_2}\)(g)

Solution

To determine which sample contains the largest number of atoms, we compare the number of moles of atoms present in each case. Since the number of atoms is directly proportional to the number of moles, the substance with the smallest molar mass per atom will contain the maximum number of atoms for the same given mass.

For \(1\,\text{g}\) of gold,

\[ \begin{aligned} \text{Moles of Au atoms} &= \frac{1}{197} = 5.08 \times 10^{-3}\,\text{mol} \end{aligned} \]

For \(1\,\text{g}\) of sodium,

\[ \begin{aligned} \text{Moles of Na atoms} &= \frac{1}{23} = 4.35 \times 10^{-2}\,\text{mol} \end{aligned} \]

For \(1\,\text{g}\) of lithium,

\[ \begin{aligned} \text{Moles of Li atoms} &= \frac{1}{7} = 1.43 \times 10^{-1}\,\text{mol} \end{aligned} \]

For \(1\,\text{g}\) of chlorine gas \(\mathrm{(Cl_2)}\), the molar mass is \(71\,\text{g mol}^{-1}\). The number of moles of chlorine molecules is therefore

\[ \begin{aligned} \text{Moles of } \mathrm{Cl_2} &= \frac{1}{71} = 1.41 \times 10^{-2}\,\text{mol} \end{aligned} \]

Since each molecule of \(\mathrm{Cl_2}\) contains two chlorine atoms, the moles of chlorine atoms are

\[ \begin{aligned} \text{Moles of Cl atoms} &= 2 \times 1.41 \times 10^{-2} = 2.82 \times 10^{-2}\,\text{mol} \end{aligned} \]

Comparing the values, lithium has the highest number of moles of atoms. Hence, \(1\,\text{g}\) of lithium contains the largest number of atoms among the given options.

Q29. Calculate the molarity of a solution of ethanol in water, in which the mole fraction of ethanol is 0.040 (assume the density of water to be one).

Solution

The mole fraction of ethanol is given as \(0.040\). This means that in the solution, for every mole of ethanol, the remaining moles correspond to water. Let us assume that the total number of moles of the solution is \(1\) mole for convenience.

\[ \begin{aligned} \text{Moles of ethanol} &= 0.040 \\ \text{Moles of water} &= 1 - 0.040 = 0.960 \end{aligned} \]

The molar mass of water is \(18\,\text{g mol}^{-1}\). Hence, the mass of water present is

\[ \begin{aligned} \text{Mass of water} &= 0.960 \times 18 \\ &= 17.28\,\text{g} \end{aligned} \]

The density of water is given as \(1\,\text{g mL}^{-1}\). Therefore, the volume of water corresponding to \(17.28\,\text{g}\) is

\[ \begin{aligned} \text{Volume of water} &= 17.28\,\text{mL} \\ &= 0.01728\,\text{L} \end{aligned} \]

Since ethanol is present in a small amount, the volume of the solution can be approximated as the volume of water. Thus, the total volume of the solution is taken as \(0.01728\,\text{L}\).

Molarity is defined as the number of moles of solute per litre of solution. Hence, the molarity of ethanol is

\[ \begin{aligned} \text{Molarity} &= \frac{0.040}{0.01728} \\ &= 2.31\,\text{mol L}^{-1} \end{aligned} \]

Therefore, the molarity of ethanol in the solution is approximately \(2.31\,\text{M}\).

Q30. What will be the mass of one \(\mathrm{^{12}C}\) atom in g?

Solution

The atomic mass of \(\mathrm{^{12}C}\) is defined as exactly \(12\,\text{u}\). Also, \(1\,\text{u}\) (atomic mass unit) is equal to \(\dfrac{1}{N_A}\) gram, where \(N_A = 6.022 \times 10^{23}\,\text{mol}^{-1}\) is Avogadro’s number.

Hence, the mass of one \(\mathrm{^{12}C}\) atom can be calculated as follows:

\[ \begin{aligned} \text{Mass of one } \mathrm{^{12}C\ atom} &= \frac{12\,\text{g}}{6.022 \times 10^{23}} \\ &= 1.99 \times 10^{-23}\,\text{g} \end{aligned} \]

Therefore, the mass of a single \(\mathrm{^{12}C}\) atom is approximately \(1.99 \times 10^{-23}\,\text{g}\).

Q31. How many significant figures should be present in the answer of the following calculations?
(i) \(\mathrm{\dfrac{0.02856\times 298.15\times 0.112}{0 5785}}\)
(ii) \(\mathrm{5 \times 5.364 }\)
(iii) \(\mathrm{0.0125 + 0.7864 + 0.0215}\)

Solution

The number of significant figures in a calculated result depends on the rules of significant figures. For multiplication and division, the answer must have the same number of significant figures as the quantity with the least significant figures. For addition, the result is limited by the least number of decimal places.

For part (i), we first count the significant figures in each term: \(\,0.02856\) has \(4\) significant figures, \(298.15\) has \(5\), \(0.112\) has \(3\), and \(0.5785\) has \(4\).

Since this expression involves multiplication and division,

\[ \begin{aligned} \text{Least significant figures} &= 3 \end{aligned} \]

Therefore, the final answer in part (i) should be reported with \(3\) significant figures.

For part (ii), the numbers involved are \(5\) and \(5.364\). The number \(5\) has only \(1\) significant figure, while \(5.364\) has \(4\).

\[ \begin{aligned} \text{Least significant figures} &= 1 \end{aligned} \]

Hence, the answer to part (ii) should contain only \(1\) significant figure.

For part (iii), the operation is addition. The numbers \(0.0125\), \(0.7864\), and \(0.0215\) all have four decimal places. Their sum is

\[ \begin{aligned} 0.0125 + 0.7864 + 0.0215 &= 0.8204 \end{aligned} \]

Since each term is given to four decimal places, the result must also be expressed to four decimal places. The value \(0.8204\) contains \(4\) significant figures.

Therefore, the answers should be reported with \(3\), \(1\), and \(4\) significant figures for parts (i), (ii), and (iii), respectively.

Q32. Use the data given in the following table to calculate the molar mass of naturally occuring argon isotopes: \[ \begin{array}{c|c} \begin{aligned} \text{Isotope}\quad&\text{Isotopic molar mass}&\text{Abundance}\\ \mathrm{^{36}Ar}\quad&\mathrm{35.96755\ g\ mol^{–1} }&\mathrm{0.337\% }\\ \mathrm{^{38}Ar}\quad&\mathrm{37.96272\ g\ mol^{–1}}&\mathrm{0.063\%}\\ \mathrm{^{40}Ar}\quad&\mathrm{39.9624\ g\ mol^{–1} }&\mathrm{99.600\%} \end{aligned} \end{array} \]

Solution

The molar mass of naturally occurring argon is obtained by taking the weighted average of the isotopic molar masses using their fractional abundances. First, the given percentage abundances are converted into fractions.

\[ \begin{aligned} \mathrm{^{36}Ar}:&\ 0.337\% = 0.00337 \\ \mathrm{^{38}Ar}:&\ 0.063\% = 0.00063 \\ \mathrm{^{40}Ar}:&\ 99.600\% = 0.99600 \end{aligned} \]

Now, the average molar mass is calculated as:

\[ \begin{aligned} \text{Molar mass of Ar} &= (35.96755 \times 0.00337) + (37.96272 \times 0.00063) + (39.9624 \times 0.99600) \\ &= 0.12121 + 0.02392 + 39.80255 \\ &= 39.94768\,\text{g mol}^{-1} \end{aligned} \]

Therefore, the molar mass of naturally occurring argon is approximately \(39.95\,\text{g mol}^{-1}\).

Q33. Calculate the number of atoms in each of the following
(i) 52 moles of Ar
(ii) 52 u of He
(iii) 52 g of He.

Solution

To calculate the number of atoms, Avogadro’s number is used, which states that \(1\,\text{mol}\) of any substance contains \(6.022 \times 10^{23}\) atoms.

For \(52\,\text{moles}\) of argon,

\[ \begin{aligned} \text{Number of atoms} &= 52 \times 6.022 \times 10^{23} \\ &= 3.13 \times 10^{25}\ \text{atoms} \end{aligned} \]

For \(52\,\text{u}\) of helium, the atomic mass of one helium atom is approximately \(4\,\text{u}\). Hence, the number of helium atoms present is obtained by dividing the total mass by the mass of one atom.

\[ \begin{aligned} \text{Number of He atoms} &= \frac{52}{4} \\ &= 13\ \text{atoms} \end{aligned} \]

For \(52\,\text{g}\) of helium, the number of moles is first calculated using its molar mass of \(4\,\text{g mol}^{-1}\).

\[ \begin{aligned} \text{Moles of He} &= \frac{52}{4} \\ &= 13\,\text{mol} \end{aligned} \]

The corresponding number of atoms is then,

\[ \begin{aligned} \text{Number of atoms} &= 13 \times 6.022 \times 10^{23} \\ &= 7.83 \times 10^{24}\ \text{atoms} \end{aligned} \]

Thus, the numbers of atoms are \(3.13 \times 10^{25}\) for \(52\,\text{mol}\) of argon, \(13\) atoms for \(52\,\text{u}\) of helium, and \(7.83 \times 10^{24}\) atoms for \(52\,\text{g}\) of helium.

Q34. A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate
(i) empirical formula,
(ii) molar mass of the gas, and
(iii) molecular formula.

Solution

The fuel gas contains only carbon and hydrogen. On combustion, all carbon appears as carbon dioxide and all hydrogen appears as water. Hence, the masses of carbon and hydrogen present in the sample are obtained from the given combustion products.

From \(3.38\,\text{g}\) of \(\mathrm{CO_2}\), the mass of carbon is calculated using the molar masses \(\mathrm{CO_2 = 44\,g\,mol^{-1}}\) and \(\mathrm{C = 12\,g\,mol^{-1}}\).

\[ \begin{aligned} \text{Mass of C} &= 3.38 \times \frac{12}{44} \\ &= 0.923\,\text{g} \end{aligned} \]

From \(0.690\,\text{g}\) of water, the mass of hydrogen is obtained using \(\mathrm{H_2O = 18\,g\,mol^{-1}}\) and \(\mathrm{H = 2\,g\,mol^{-1}}\).

\[ \begin{aligned} \text{Mass of H} &= 0.690 \times \frac{2}{18} \\ &= 0.0767\,\text{g} \end{aligned} \]

Now, the number of moles of each element is calculated.

\[ \begin{aligned} \text{Moles of C} &= \frac{0.923}{12} = 0.0769 \\ \text{Moles of H} &= \frac{0.0767}{1} = 0.0767 \end{aligned} \]

Dividing by the smaller value gives an approximate mole ratio of

\[ \begin{aligned} \text{C} &: \frac{0.0769}{0.0767} \approx 1 \\ \text{H} &: \frac{0.0767}{0.0767} = 1 \end{aligned} \]

Hence, the empirical formula of the gas is \(\mathrm{CH}\).

Next, the molar mass of the gas is calculated using the STP data. At STP, \(22.4\,\text{L}\) of any gas corresponds to one mole. Given that \(10.0\,\text{L}\) of the gas has a mass of \(11.6\,\text{g}\),

\[ \begin{aligned} \text{Molar mass} &= \frac{11.6}{10.0} \times 22.4 \\ &= 26.0\,\text{g mol}^{-1} \end{aligned} \]

Thus, the molar mass of the gas is approximately \(26\,\text{g mol}^{-1}\).

The empirical formula mass of \(\mathrm{CH}\) is

\[ \begin{aligned} 12 + 1 = 13\,\text{g mol}^{-1} \end{aligned} \]

The ratio of molar mass to empirical formula mass is

\[ \begin{aligned} \frac{26}{13} = 2 \end{aligned} \]

Therefore, the molecular formula is obtained by multiplying the empirical formula by \(2\), giving \(\mathrm{C_2H_2}\).

Hence, the empirical formula is \(\mathrm{CH}\), the molar mass is about \(26\,\text{g mol}^{-1}\), and the molecular formula of the welding gas is \(\mathrm{C_2H_2}\).

Q35. Calcium carbonate reacts with aqueous HCl to give \(\mathrm{CaCl_2}\) and CO2 according to the reaction, \[\mathrm{CaCO_3 (s) + 2 HCl (aq) → CaCl_2 (aq) + CO_2(g) + H_2O(l)}\] What mass of \(\mathrm{CaCO_3}\) is required to react completely with 25 mL of 0.75 M HCl?

Solution

The balanced chemical equation for the reaction is \(\mathrm{CaCO_3 + 2HCl \rightarrow CaCl_2 + CO_2 + H_2O}\). From this equation, \(1\) mole of calcium carbonate reacts with \(2\) moles of hydrochloric acid.

First, the number of moles of \(\mathrm{HCl}\) present in \(25\,\text{mL}\) of a \(0.75\,\text{M}\) solution is calculated. The volume is converted to litres as \(25\,\text{mL} = 0.025\,\text{L}\).

\[ \begin{aligned} \text{Moles of HCl} &= 0.75 \times 0.025 \\ &= 0.01875\,\text{mol} \end{aligned} \]

According to the stoichiometry, the moles of \(\mathrm{CaCO_3}\) required are half the moles of \(\mathrm{HCl}\):

\[ \begin{aligned} \text{Moles of } \mathrm{CaCO_3} &= \frac{0.01875}{2} \\ &= 0.009375\,\text{mol} \end{aligned} \]

Now, using the molar mass of calcium carbonate, \(\mathrm{CaCO_3 = 100\,g\,mol^{-1}}\),

\[ \begin{aligned} \text{Mass of } \mathrm{CaCO_3} &= 0.009375 \times 100 \\ &= 0.9375\,\text{g} \end{aligned} \]

Therefore, approximately \(0.94\,\text{g}\) of calcium carbonate is required to react completely with \(25\,\text{mL}\) of \(0.75\,\text{M}\) hydrochloric acid.

Q36. Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with aqueous hydrochloric acid according to the reaction \[\mathrm{4 HCl (aq) + MnO_2(s) → 2H_2O (l) + MnCl_2(aq) + Cl_2 (g)}\] How many grams of HCl react with 5.0 g of manganese dioxide?

Solution

The balanced chemical equation for the reaction is \(\mathrm{4HCl + MnO_2 \rightarrow 2H_2O + MnCl_2 + Cl_2}\). From this equation, \(1\) mole of manganese dioxide reacts with \(4\) moles of hydrochloric acid.

First, the number of moles of \(\mathrm{MnO_2}\) present in \(5.0\,\text{g}\) is calculated. The molar mass of \(\mathrm{MnO_2}\) is

\[ \begin{aligned} \text{Molar mass of } \mathrm{MnO_2} &= 55 + 2(16) = 87\,\text{g mol}^{-1} \end{aligned} \]

\[ \begin{aligned} \text{Moles of } \mathrm{MnO_2} &= \frac{5.0}{87} = 5.75 \times 10^{-2}\,\text{mol} \end{aligned} \]

According to the stoichiometry of the reaction, the moles of \(\mathrm{HCl}\) required are:

\[ \begin{aligned} \text{Moles of HCl} &= 4 \times 5.75 \times 10^{-2} \\ &= 0.230\,\text{mol} \end{aligned} \]

Now, using the molar mass of hydrochloric acid, \(\mathrm{HCl = 36.5\,g\,mol^{-1}}\), the mass of \(\mathrm{HCl}\) consumed is:

\[ \begin{aligned} \text{Mass of HCl} &= 0.230 \times 36.5 \\ &= 8.40\,\text{g} \end{aligned} \]

Therefore, approximately \(8.4\,\text{g}\) of hydrochloric acid react with \(5.0\,\text{g}\) of manganese dioxide.