$\mathrm{m}=0.25 \mathrm{g}, \mathrm{V}_{1}=40 \mathrm{ml}, \mathrm{T}_{1}=300 \mathrm{K}, \mathrm{P}_{1}=725 \mathrm{mm}-25 \mathrm{mm}=700 \mathrm{mm}$
$\mathrm{P}_{0}=760 \mathrm{mm}, \mathrm{T}_{0}=273 \mathrm{K}, \mathrm{V}_{0}=?$
$\mathrm{V}_{0}=\frac{\mathrm{P}_{1} \mathrm{V}_{1}}{\mathrm{T}_{1}} \times \frac{\mathrm{T}_{\mathrm{o}}}{\mathrm{P}_{0}}=\frac{700 \mathrm{mm} \times 40 \mathrm{ml}}{300 \mathrm{K}} \times \frac{273 \mathrm{K}}{760 \mathrm{mm}}=33.53 \mathrm{ml}$
At STP, 22400 ml of nitrogen occupies 22400 ml.
22400mL OF $\mathrm{N}_{2}$ at STP weighs $=28 \mathrm{gm}$
Hence the mass of nitrogen which corresponds to $33.53 \mathrm{ml}$ is $\frac{28 \mathrm{g} \times 33.53 \mathrm{ml}}{22400 \mathrm{ml}}=0.0419 \mathrm{g}$.
The percentage of nitrogen is $0.0419 \mathrm{g} \times \frac{100}{0.25 \mathrm{g}}=16.76 \%$
Hence, the correct option is B.
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$\mathrm{NH}_3, \mathrm{SO}_2, \mathrm{SiO}_2, \mathrm{BeCl}_2, \mathrm{CO}_2, \mathrm{H}_2 \mathrm{O}, \mathrm{CH}_4, \mathrm{BF}_3$
$kJ\,mol ^{-1}$ (Nearest integer)
Given : $\Delta_{ dis } H _{ Cl _{2(g)}}^{\circ}=240\,kJ\,mol ^{-1}$.
$\Delta_{ eg } H _{ Cl _{(g)}}^{\circ}=-350\,kJ\,mol ^{-1}$,
$\Delta_{ hyd } H _{ Cl i _{( j )}^{\circ}}^{\circ}=-380\,kJ\,mol ^{-1}$