Question

If $4g$ of oxygen diffuse through a very narrow hole, how much hydrogen, would have diffused under identical conditions?

• A. $16g$
• B. $1g$
• C. $1/4g$
• D. $64g$

Answer 1

Answer: (B)

$1g$

Time taken by the $4g$ of $O_2$ to diffuse $=1$

In the same time $\frac{massof{H}_{2}diffused}{timetaken}$

Rrate of diffusion of $H_2,{rH}_2=\frac{V_{H_2}}{t}$ -----------(1)

Rate of diffusion of $O_2,{rO}_2=\frac{V_{O_2}}{t}$ ----------(2)

On dividing equation (1) and (2), we get,

$\frac{r_{H_2}}{r_{O_2}}=\frac{V_{H_2}}{V_{O_2}}$

$\frac{r_{H_2}}{r_{O_2}}=\frac{\frac{m_{H_2}}{d_{H_2}}}{\frac{m_{o_2}}{d_{o_2}}}$

$\frac{r_{H_2}}{r_{O_2}}=\frac{\frac{m_{H_2}}{0.08986}}{\frac{4}{1.429}}$

$\frac{r_{H_2}}{r_{O_2}}=3.975\times m_{H_2}$

According to Grahams law of diffusion

$m_{H_2}=1.006g$

$\frac{r_{H_2}}{r_{O_2}}=\sqrt{\frac{M_{O_2}}{M_{H_2}}}$

$\frac{r_{H_2}}{r_{O_2}}=\sqrt{\frac{32}{2}}=4$

Combining equation(3) and (4)

$\frac{r_{H_2}}{r_{O_2}}=3.975\times m_{H_2}=4$

$m_{H_2}=1.006g$