Question

$0.3\text{g}$ of an organic compound gives $50\text{mL}$ of $N_2$ gas at ${27}^{\circ }\text{C}$ and $715\text{mm}$ of Hg pressure. Calculate the percentage of $N_2$, given that the aqueous tension at ${27}^{\circ }\text{C}$ is $15\text{mm}$ of $Hg$.

• A. $17.5\%$
• B. $41.9\%$
• C. $1.75\%$
• D. $4.19\%$

Answer 1

The correct option is A $17.5\%$

% of $N_2$ by Duma's method = $\frac{28\times V_{N_2}\times 100}{22400\times massoforganiccompound}$
Now, calculating the vol of $N_2$ at STP using ;
$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$
$\frac{P_{N_2}\times V_{N_2}}{T{}_{{27}^{0}C}}=\frac{P_{N_2}\times V_{N_2}}{T{}_{0^{0}C}}-----\left(1\right)$
$P_{N_2}=(P_{moistgas}-aq.tension)$
= 715 - 15 = 700 mm of Hg
putting the value of $P_{N_2}$ in equation (1)
$\frac{700\times 50}{300}=\frac{760\times V_{N_2}}{273}$

$V_{N_2}=\frac{7\times 50\times 273}{760\times 3}$
$\therefore$% of $N_2=\frac{28\times 41.91\times 100}{22400\times 0.3}$
=17.5 %