Chemistry - Concentration Terms Question with Solution | TestHub

In 2g sample$\to$CO₂Liberated

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$\equiv 1\mathrm{0 }\text{mM}$of NaHCO₃in mixture

Moles of NaHCO₃in 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$\equiv \text{15}\times \frac{2}{1·5}=\text{20 mM}·$

20 mM HCl will be utilised by

10 mM${\text{NaHCO}}_3{+ \text{5 mM Na}}_2{\text{CO}}_{\mathrm{3 }}\text{present in 2g sample}$

${\text{NaHCO}}_3=\text{10 mM}=\text{10}\times 10^{-3}\times 84=\text{0.84g }\left(\text{42\%}\right)$

${\text{Na}}_2{\text{CO}}_3=\text{5 mM}=5\times 10^{-3}\times 106=\text{0.53g}=\left(\text{26.5\%}\right)$

${\text{Na}}_2{\text{SO}}_4= \text{(By difference)} =\text{0.63g }\left(\text{31.5\%}\right)$