A glass bulb contains 2.24 L of $H_2$ and 1.12 L of ${\mathrm{D}}_2$ at STP. It is connected a fully evacuated bulb by a stop-cock with a small opening. The stop-cock is opened for sometime and then closed. The first bulb now contains 0.10 g of ${\mathrm{H}}_2$. The percentage of ${\mathrm{H}}_2$ in the mixture is

$\begin{array}{l}1\mathrm{mole}\mathrm{of}{\mathrm{H}}_2\left(2\mathrm{g}\right)\mathrm{occupies}22.4\mathrm{L}\\ 2.24\mathrm{L}\mathrm{of}{\mathrm{H}}_2\mathrm{at}\mathrm{STP}=\frac{2}{22.4}\times 2.24=0.2\mathrm{g}\\ 1.12\mathrm{L}\mathrm{of}{\mathrm{D}}_2\left(4\mathrm{g}\right)\mathrm{at}\mathrm{STP}=\frac{4}{22.4}\times 1.12=0.2\mathrm{g}\end{array}$

Mass of $H_2$ diffused in given time = 0.2 g - 0.1 g = 0.1 g

$\frac{{\mathrm{m}}_{{\mathrm{H}}_2}}{{\mathrm{m}}_{{\mathrm{D}}_2}}=\mathrm{\hspace{0.17em}}\sqrt{\frac{{\mathrm{M}}_{{\mathrm{H}}_2}}{{\mathrm{M}}_{{\mathrm{D}}_2}}}$

$\begin{array}{l}{\mathrm{m}}_{{\mathrm{D}}_2}=\sqrt{\frac{{\mathrm{M}}_{{\mathrm{D}}_2}}{{\mathrm{M}}_{{\mathrm{H}}_2}}}\times {\mathrm{m}}_{{\mathrm{H}}_2}=\sqrt{\frac{4}{2}}\times 0.1\mathrm{g}=0.14\mathrm{g}\\ \mathrm{\%}\mathrm{of}{\mathrm{H}}_2\mathrm{in}\mathrm{this}\mathrm{bulb}=\frac{0.1}{0.1+0.14}\times 100=41.6\mathrm{\%}\end{array}$