Chemistry - Mole Concept Question with Solution | TestHub

${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$is 86% by weight
So weight of${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$= 86 g
Weight of solution = 100 g
So volume of solution$=\frac{100}{1.787}\mathrm{m}\mathrm{L}=\frac{100}{1.787\times 1000}\mathrm{L}$
${\mathrm{M}}_{{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4}=\frac{86}{98\times \frac{100}{1.787\times 1000}}=15.68$
Let V mL of this${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$are used to prepare 1 L of 0.2 M${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$.
So mmoles of concentrated${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4=$mmoles of dilute${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$
So$\mathrm{V}\times 15.68=1000\times 0.2$
So V = 12.75 mL.