c
$\underset{\substack{\text { Mol..wt.45g }}}{\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{NH}_2} \xrightarrow{\mathrm{N}_2 \mathrm{NO}_2+\mathrm{HCl}} \xrightarrow{\mathrm{H}_2 \mathrm{O}} \mathrm{CH}_3 \mathrm{CH}_2-\mathrm{OH}+\underset{14 \mathrm{~g}}{\mathrm{~N}_2}$
given: $\mathrm{N}_2$ evolved is $2.24 \mathrm{~L}$ i.e. $0.1 \mathrm{~mole}$.
1.e. $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{NH}_2$ (ethyl amine) will be $4.5 \mathrm{~g}$
$(=0.1 \mathrm{~mole})$
Hence the answer $=45 \times 10^{-1} \mathrm{~g}$