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

In 2g sample $ \rightarrow $ CO₂ Liberated

Moles of CO₂ $= \left(\frac{7 5 0}{7 6 0} \times \frac{1 2 3 \text{.} 9}{1 0 0 0} \times \frac{1}{0 \text{.} 0 8 2} \times \frac{1}{2 9 8}\right) = 5 \times 1 0^{- 3}$

moles = 5 mM $\equiv 1 0 \, \text{mM}$ of NaHCO₃ in mixture

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

1.5g sample $\equiv \text{150} \times \frac{1}{1 0} = \text{15 mM}$ 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

$10\, mM\, NaHCO _{3}+5 mM Na _{2} CO _{3}$ present in $2 g$ sample

$NaHCO _{3}=10 mM =10 \times 10^{-3} \times 84=0.84 g (42 \%)$

$Na _{2} CO _{3}=5 mM =5 \times 10^{-3} \times 106=0.53 g =(26.5 \%)$

$Na _{2} SO _{4}=($ By difference $)=0.63 g (31.5 \%)$