Chemistry - Concentration Terms Question with Solution | TestHub

Assume that purity of KI is $\mathrm{x}\%$

then pure mass of $\mathrm{KI}=\frac{10\mathrm{x}}{100}=0.1\mathrm{xgm}$

mol of $\mathrm{KI}=\frac{0.1\mathrm{x}}{166}\mathrm{mol}$

and produced mol of ${\mathrm{I}}_2=\frac{0.1\mathrm{x}}{166}\times \frac{4}{8}$

we know

Mol of ${\mathrm{I}}_2=[\mathrm{Mol}$ of $\mathrm{NaI}\times \frac{1}{2}+\mathrm{Mol}$ of ${\mathrm{Cl}}_2]$ $\Rightarrow \frac{0.1}{166}\times \frac{4}{8}=\left[\frac{0.3}{150}\times \frac{1}{2}+\frac{0.497}{71}\right]$

$\Rightarrow \frac{0.1x}{166\times 2}=0.001+0.007=0.008$

$x=26.56$ Ans.

$\&\mathrm{Mol}$ of ${\mathrm{I}}_2=0.008$ Ans.

Mol of ${\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3=\mathrm{Mol}$ of NaI

$\frac{20}{1000}\times {\mathrm{M}}_{{\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3}=0.002$

${\mathrm{M}}_{{\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3}=0.1\mathrm{mol}/\mathrm{lit}$.

and we know,

$\frac{5}{8}$ [Reacted mol of KI in 1st reaction] $=$ [Given mol of ${\mathrm{H}}_2{\mathrm{SO}}_4-$ Reacted mol of ${\mathrm{H}}_2{\mathrm{SO}}_4$ in $2^{\text{nd }}$ reaction] Assume $\%[\mathrm{w}/\mathrm{w}]$ of ${\mathrm{H}}_2{\mathrm{SO}}_4$ is w , then

$\left[\frac{5}{8}\times 0.016\right]=\left[10\times 1.47\times \frac{w}{100}\times \frac{1}{98}-\frac{1}{2}\times \frac{20}{100}\times 8\times \frac{1}{40}\right]$

$[0.01]=\left[\frac{0.147}{98}\mathrm{w}-0.02\right]$

$\frac{0.147}{98}\mathrm{w}=0.03$

$\mathrm{w}=20\%$