Question

A mixture of $NaI$ and $NaCl$ on reaction with $H_{2}S{O}_4$ gave $N{a}_{2}S{O}_4$ equal to weight of original mixture taken. The percentage of $NaI$ in the mixture is

• A. $82.38$
• B. $26.38$
• C. $62.38$
• D. $28.38$

Answer 1

The correct option is B $26.38$

Mixture of $NaI$ and $NaCl$

Let $NaI$ be $xgm$

$NaCl$ be $ygm$

$2NaI+H_{2}S{O}_4\to N{a}_{2}S{O}_4+2HI$

$2NaCl+H_{2}S{O}_4\to N{a}_{2}S{O}_4+2HCl$

$MNaI=1.50gm,MNaCl=58.5gm,MN{a}_{2}S{O}_4=142gm$

$2$ mole $NaI$ required to give $1$ mole $N{a}_{2}S{O}_4$ same for $NaCl:2$ mole give $1$ mole $N{a}_{2}S{O}_4$.

Now mass of $N{a}_{2}S{O}_4=wt-$ of original mixture $=(x+y1gm)$

$nN{a}_{2}S{O}_4=(\frac{x}{150}+\frac{y}{58.5})\times \frac{1}{2}$

$\therefore$ Mass of $N{a}_{2}S{O}_4=(\frac{x}{150}+\frac{y}{58.5})\times \frac{1}{2}\times 142=x+y$

$\frac{71x}{150}+\frac{71y}{58.5}=x+y$

$0.473x+1.21y=x+y$

$\Rightarrow 0.52x=0.21y$

$\therefore y$ of $NaI=\frac{x}{x+y}\times 100$

$\frac{x}{x+\frac{0.527}{0.21}x}\times 100=\frac{0.21}{0.737}\times 100$

$=28.4gm$