Question

If the percentage of nitrogen in an organic compound is $12.5\%$ then the mass of organic compound taken so as to produce $50\text{mL}$ of $N_2$ at $300\text{K}$ and $684\text{mm Hg}$ pressure is $x\text{g}$. The value of $\frac{50}{3}x$ is:
$(R=0.08\text{atm-L/mol-K})$

• A. 7.00
• B. 7
• C. 7.0

Answer 1

Answer: (A), (B), (C)

Given;
$P=684\text{mm Hg}=684/760\text{atm}$
$T=300\text{K}$
$R=0.08\text{atm-L/mol-K}$
Molecular mass of nitrogen $=28\text{g/mol}$
From, $PV=nRT$
$\text{Weight of Nitrogen in organic compound}=\frac{PV\times \text{M.wt}}{RT}$
$=\frac{684\times 50\times {10}^{-3}\times 28}{0.08\times 760\times 300}=0.0525\text{g}$
$\text{\% of N in organic comp.}=\frac{\text{Mass of N in organic comp}}{\text{Total mass of organic comp.(w)}}\times 100$
$\Rightarrow 12.5=\frac{0.0525}{w}\times 100$
$\Rightarrow w=0.42$
So, $\frac{50}{3}x=7$