Some Basic Concepts of Chemistry 2 Question 40
40. A $2.0 \mathrm{~g}$ sample of a mixture containing sodium carbonate, sodium bicarbonate and sodium sulphate is gently heated till the evolution of $\mathrm{CO}{2}$ ceases. The volume of $\mathrm{CO}{2}$ at $750 \mathrm{~mm}$ $\mathrm{Hg}$ pressure and at $298 \mathrm{~K}$ is measured to be $123.9 \mathrm{~mL}$. A $1.5 \mathrm{~g}$ of the same sample requires $150 \mathrm{~mL}$ of $(\mathrm{M} / 10) \mathrm{HCl}$ for complete neutralisation. Calculate the percentage composition of the components of the mixture. $(1992,5 \mathrm{M})$
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Solution:
- $\mathrm{CO}_{2}$ is evolved due to following reaction :
$$ 2 \mathrm{NaHCO}{3} \longrightarrow \mathrm{Na}{2} \mathrm{CO}{3}+\mathrm{H}{2} \mathrm{O}+\mathrm{CO}_{2} $$
Moles of $\mathrm{CO}_{2}$ produced $=\frac{p V}{R T}$
$$ \begin{aligned} & =\frac{750}{760} \times \frac{123.9}{1000} \times \frac{1}{0.082 \times 298} \ & =5 \times 10^{-3} \end{aligned} $$
$\Rightarrow$ Moles of $\mathrm{NaHCO}_{3}$ in $2 \mathrm{~g}$ sample $=2 \times 5 \times 10^{-3}=0.01$
$\Rightarrow$ millimol of $\mathrm{NaHCO}_{3}$ in $1.5 \mathrm{~g}$ sample
$$ =\frac{0.01}{2} \times 1.5 \times 1000=7.5 $$
Let the $1.5 \mathrm{~g}$ sample contain $x$ millimol $\mathrm{Na}{2} \mathrm{CO}{3}$, then
$$ 2 x+7.5=\text { millimol of } \mathrm{HCl}=15 $$
$$ \Rightarrow \quad x=3.75 $$
$$ \begin{aligned} \Rightarrow \quad \text { Mass of } \mathrm{NaHCO}{3} & =\frac{7.5 \times 84}{1000}=0.63 \mathrm{~g} \ \text { Mass of } \mathrm{Na}{2} \mathrm{CO}{3} & =\frac{3.75 \times 106}{1000}=0.3975 \mathrm{~g} \ \Rightarrow % \text { mass of } \mathrm{NaHCO}{3} & =\frac{0.63}{1.50} \times 100=42 % \ % \text { mass of } \mathrm{Na}{2} \mathrm{CO}{3} & =\frac{0.3975}{1.5} \times 100 \ & =26.5 % \end{aligned} $$
