JEE Main 2019 — Hydrogen Question with Solution

Molarity of ${\mathrm{H}}_2\mathrm{O}=\frac{\mathrm{Volume}\mathrm{\hspace{0.17em}}\mathrm{strength}}{11.2}$

$\mathrm{M}=\frac{11.2}{11.2}=1$

$\therefore$ $1$ molar means $1000\mathrm{ }\mathrm{mL}$ solution contains $1\hspace{0.17em}\mathrm{mol}\hspace{0.17em}{\mathrm{H}}_2{\mathrm{O}}_2=1\times 34\hspace{0.17em}\mathrm{g}$

Since, the strength is represented as weight of the solute in $1 \mathrm{L}$ of the solution. The strength of the solution will be $34 \mathrm{g} {\mathrm{L}}^{-1}$.

Note: The question should've mentioned that we need to find the mass by volume percentage.

Percentage w/v $=\frac{34}{1000}\times 100=3.4\mathrm{\%}$