Question

A $2.0g$ sample of a mixture containing $Na_{2}CO_{3},NaHCO_{3}andNa_{2}SO_{4}$ is gently heated till the evolution of $CO_{2}$ stops. The evolved $CO_{2}$ has volume of $123.9mL$ measured at $750mmHg$ pressure and $298K$ .
$1.5g$ of the same sample is completely neutralised by $150mL$ of $0.1MHCl$ . The percentage composition of $\text{Na}_{2} \text{SO}_{4}$ in the original mixture is:

Answer 1

In 2g sample $→$ CO₂ Liberated
Moles of CO₂ $= \left(\frac{7 5 0}{7 6 0} \times \frac{1 2 3 \text{.} 9}{1 0 0 0} \times \frac{1}{0 \text{.} 0 8 2} \times \frac{1}{2 9 8}\right) = 5 \times 1 0^{- 3}$
= 5 x 10^-3
moles = 5 mM $≡ 1 0 \, \text{mM}$ of NaHCO₃ in mixture
Moles of NaHCO₃ in 2g sample = 2 x 5 x 10 mM
1.5g sample $≡ \text{150} \times \frac{1}{1 0} = \text{15 mM}$ of HCl
2.0g sample $≡ \text{15} \times \frac{2}{1 · 5} = \text{20 mM} ·$
20 mM HCl will be utilised by
10 mM $\text{NaHCO}_{3} + \, \text{5 mM Na}_{2} \text{CO}_{3 \, } \text{present in 2g sample}$
$NaHCO _3=10 mM =10 \times 10^{-3} \times 84=0.84 g (42 \%)$
$Na _2 CO _3=5 mM =5 \times 10^{-3} \times 104=0.52 g =(26 \%)$
$ Na _2 SO _4==0.64 g (32 \%) $