Question

A 2.0 g sample of a mixture containing sodium carbonate, sodium bicarbonate and sodium sulphate is gently heated till the evolution of CO₂ ceases. The volume of CO₂ 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 (M/10) HCl for complete neutralization. Calculate the percentage composition of ${\text{Na}}_2{\text{SO}}_4$ in the original mixture.

Answer 1

Answer: 31.50

In 2g sample $\to$ CO₂ Liberated

Moles of CO₂ $=\left(\frac{750}{760}\times \frac{123\text{.}9}{1000}\times \frac{1}{0\text{.}082}\times \frac{1}{298}\right)=5\times 10^{-3}$

moles = 5 mM $\equiv 10\text{mM}$ of NaHCO₃ in mixture

Moles of NaHCO₃ in 2g sample = 2 x 5 x 10 mM

1.5g sample $\equiv \text{150}\times \frac{1}{10}=\text{15 mM}$ of HCl

2.0g sample $\equiv \text{15}\times \frac{2}{1\cdot 5}=\text{20 mM}\cdot$

20 mM HCl will be utilised by

$10mMNaHC{O}_3+5mMN{a}_{2}C{O}_3$ present in $2g$ sample

$NaHC{O}_3=10mM=10\times 10^{-3}\times 84=0.84g(42\%)$

$N{a}_{2}C{O}_3=5mM=5\times 10^{-3}\times 106=0.53g=(26.5\%)$

$N{a}_{2}S{O}_4=($ By difference $)=0.63g(31.5\%)$