JEE Advance - Chemistry (1992 - No. 3)
A 2.0 g sample of a mixture containing sodium carbonate, sodium bicarbonate and sodium suplphate is gently heated till the evolution of CO2 ceases. The volume of CO2 at 750 mm Hg pressure and at 298K is measured to be 123.9 ml. A 1.5g of the same sample requires 150 ml. of (M/10) HCl for complete neutralisation. Calculate the % composition of the components of the mixture.
Na₂CO₃ - 26.5%, NaHCO₃ - 42%, Na₂SO₄ - 31.5%
Na₂CO₃ - 30%, NaHCO₃ - 40%, Na₂SO₄ - 30%
Na₂CO₃ - 25%, NaHCO₃ - 45%, Na₂SO₄ - 30%
Na₂CO₃ - 20%, NaHCO₃ - 50%, Na₂SO₄ - 30%
Na₂CO₃ - 31.5%, NaHCO₃ - 26.5%, Na₂SO₄ - 42%
Explanation
$\mathrm{CO}_2$ is evolved due to following reaction :
$$ 2 \mathrm{NaHCO}_3 \longrightarrow \mathrm{Na}_2 \mathrm{CO}_3+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2 $$
$$ \begin{aligned} \text { Moles of } \mathrm{CO}_2 \text { produced } & =\frac{p V}{R T}=\frac{750}{760} \times \frac{123.9}{1000} \times \frac{1}{0.082 \times 298} \\\\ & =5 \times 10^{-3} \end{aligned} $$
$\begin{aligned} & \Rightarrow \text { Moles of } \mathrm{NaHCO}_3 \text { in } 2 \mathrm{~g} \text { sample }=2 \times 5 \times 10^{-3}=0.01 \\\\ & \Rightarrow \text { millimol of } \mathrm{NaHCO}_3 \text { in } 1.5 \mathrm{~g} \text { sample } \\\\ & \qquad=\frac{0.01}{2} \times 1.5 \times 1000=7.5\end{aligned}$
Let the $1.5 \mathrm{~g}$ sample contain $x$ millimol $\mathrm{Na}_2 \mathrm{CO}_3$, then
$$ \begin{array}{rlrl} & 2 x+7.5 =\text { millimol of } \mathrm{HCl}=15 \\\\ & \Rightarrow x =3.75 \end{array} $$
$\begin{aligned} \Rightarrow \text { Mass of } \mathrm{NaHCO}_3 =\frac{7.5 \times 84}{1000}=0.63 \mathrm{~g} \\\\ \text { Mass of } \mathrm{Na}_2 \mathrm{CO}_3=\frac{3.75 \times 106}{1000}=0.3975 \mathrm{~g}\end{aligned}$
$\begin{aligned} \Rightarrow \% \text { mass of } \mathrm{NaHCO}_3 =\frac{0.63}{1.50} \times 100=42 \% \\\\ \% \text { mass of } \mathrm{Na}_2 \mathrm{CO}_3 =\frac{0.3975}{1.5} \times 100=26.5 \%\end{aligned}$
$\begin{aligned} \% \text { of } \mathrm{Na}_2 \mathrm{SO}_4 \text { in the sample } & =100-(42+26.5) \\\\ & =100-68.5=31.5 \%\end{aligned}$
$$ 2 \mathrm{NaHCO}_3 \longrightarrow \mathrm{Na}_2 \mathrm{CO}_3+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2 $$
$$ \begin{aligned} \text { Moles of } \mathrm{CO}_2 \text { produced } & =\frac{p V}{R T}=\frac{750}{760} \times \frac{123.9}{1000} \times \frac{1}{0.082 \times 298} \\\\ & =5 \times 10^{-3} \end{aligned} $$
$\begin{aligned} & \Rightarrow \text { Moles of } \mathrm{NaHCO}_3 \text { in } 2 \mathrm{~g} \text { sample }=2 \times 5 \times 10^{-3}=0.01 \\\\ & \Rightarrow \text { millimol of } \mathrm{NaHCO}_3 \text { in } 1.5 \mathrm{~g} \text { sample } \\\\ & \qquad=\frac{0.01}{2} \times 1.5 \times 1000=7.5\end{aligned}$
Let the $1.5 \mathrm{~g}$ sample contain $x$ millimol $\mathrm{Na}_2 \mathrm{CO}_3$, then
$$ \begin{array}{rlrl} & 2 x+7.5 =\text { millimol of } \mathrm{HCl}=15 \\\\ & \Rightarrow x =3.75 \end{array} $$
$\begin{aligned} \Rightarrow \text { Mass of } \mathrm{NaHCO}_3 =\frac{7.5 \times 84}{1000}=0.63 \mathrm{~g} \\\\ \text { Mass of } \mathrm{Na}_2 \mathrm{CO}_3=\frac{3.75 \times 106}{1000}=0.3975 \mathrm{~g}\end{aligned}$
$\begin{aligned} \Rightarrow \% \text { mass of } \mathrm{NaHCO}_3 =\frac{0.63}{1.50} \times 100=42 \% \\\\ \% \text { mass of } \mathrm{Na}_2 \mathrm{CO}_3 =\frac{0.3975}{1.5} \times 100=26.5 \%\end{aligned}$
$\begin{aligned} \% \text { of } \mathrm{Na}_2 \mathrm{SO}_4 \text { in the sample } & =100-(42+26.5) \\\\ & =100-68.5=31.5 \%\end{aligned}$
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