Question

In Duma's method for estimation of nitrogen , 0.25 g of an organic compound gave 40 mL of nitogen collected at 300 K temperature and 725 mm pressure . If the aqueous tension at 300 K is 25 mm , the percentage of nitrogen in the compound is :

• A. 17.36
• B. 15.76
• C. 16.76
• D. 18.20

Answer 1

The correct option is C 16.76

Given :

$m=0.25g,V_1=40ml,T_1=300K,p_1=725mm-25mm=700mm.$

$P_0=760mm,T_0=273K,V_0=?$

$V_0=\frac{P_1{V}_1}{T_1}\times \frac{T_o}{P_0}=\frac{700mm\times 40ml}{300K}\times \frac{273K}{760mm}=33.53ml.$

At $\text{STP}$ $22400ml$ of nitrogen occupies $22400\text{ml}$

$\text{22400 mL OF}{N}_{2}at\text{STP weighs=28 gm}$

Hence the mass of nitrogen which which corrresponds to $\text{33.53 ml}$ is $\frac{28g\times 33.53ml}{22400ml}=0.049g$.

The percentage of nitrogen is $0.0419g\times \frac{100}{0.25g}=16.76$%.

Hence, the correct option is A.