Some Basic Concepts of Chemistry - Result Question 39
40. A 2.0 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 mm Hg pressure and at 298 K is measured to be 123.9 mL . A 1.5 g of the same sample requires 150 mL of $(\mathrm{M} / 10) \mathrm{HCl}$ for complete neutralisation. Calculate the percentage composition of the components of the mixture. (1992, 5M)
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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 g sample $=2 \times 5 \times 10^{-3}=0.01$
$\Rightarrow$ millimol of $\mathrm{NaHCO}_3$ in 1.5 g sample
$=\frac{0.01}{2} \times 1.5 \times 1000=7.5$
Let the 1.5 g sample contain $x$ millimol $\mathrm{Na}_2 \mathrm{CO}_3$, then
$\begin{aligned} & 2 x+7.5=\text { millimol of } \mathrm{HCl}=15 \\ & \Rightarrow \quad x=3.75 \\ \end{aligned}$
$\begin{aligned} \Rightarrow \quad \text { Mass of } \mathrm{NaHCO}_3 & =\frac{7.5 \times 84}{1000}=0.63 \mathrm{~g} \\ \text { Mass of } \mathrm{Na}_2 \mathrm{CO}_3 & =\frac{3.75 \times 106}{1000}=0.3975 \mathrm{~g} \\ \Rightarrow % \text { mass of } \mathrm{NaHCO}_3 & =\frac{0.63}{1.50} \times 100=42 % \\ % \text { mass of } \mathrm{Na}_2 \mathrm{CO}_3 & =\frac{0.3975}{1.5} \times 100 \\ & =26.5 % \end{aligned}$
BETA
