Question

The rate of diffusion of a gas having molecular weight just double of nitrogen gas is $56$ ml per sec. The rate of diffusion of nitrogen gas will be:

• A. $79.19ml/sec$
• B. $112ml/sec$
• C. $56ml/sec$
• D. $90ml/sec$

Answer 1

Answer: (A)

$79.19ml/sec$

Mass of nitrogen gas $\left(m_1\right)=M$

Mass of another gas $\left(m_2\right)=2M$

Rate of diffusion of nitrogen gas $\left(r_1\right)=r=?$

Rate of diffusion of another gas $\left(r_2\right)=56mL/sec$

Graham's law states that the rate of diffusion or of effusion of a gas is inversely proportional to the square root of its molecular weight.

$r\propto \frac{1}{\sqrt{m}}$

$\frac{r_1}{r_2}=\sqrt{\frac{m_2}{m_2}}$

$\Rightarrow \frac{r}{56}=\sqrt{\frac{2M}{M}}$

$\Rightarrow r=56\sqrt{2}$

$\Rightarrow r=79.19mL/sec$

Hence, rate of diffusion of nitrogen gas will be $79.19mL/sec$.