Question

The time taken for a certain volume of gas to diffuse through a small hole was 2 minute .under similar conditions an equal volume of oxygen took 5.65 minute to pass .the molecular formula of the gas is

Answer 1

Answer: The molecular formula of the gas is $C_{13}H_{21}NO_2S$

Explanation:

Rate of effusion is defined as the volume of gas effused in a given time 't'.

Formula used to calculate the rate of effusion is:

$Rate=\frac{Volume}{Time}$

According to Graham's Law, the rate of effusion is directly proportional to the square root of the mass of the gas.

$\text{Rate of effusion}\propto \frac{1}{\sqrt{\text{Mass of gas}}}$

Combining two equations, we get:

$\left(\frac{\frac{V}{t_1}}{\frac{V}{t_2}}\right)=\sqrt{\frac{M_1}{M_2}}$

where,

$M_1$ = molar mass of unknown gas  = ?

$M_2$ = molar mass of $O_2$ gas  = 32 amu

$t_1$ = time of effusion of unknown gas  = 2 min

$t_2$ = time of effusion of $O_2$ gas = 5.65 min

We are provided that same volume of gases are effused, so the molar mass of unknown gas is calculated from above equation, we get:

$\left(\frac{\frac{V}{2}}{\frac{V}{5.65}}\right)=\sqrt{\frac{M_1}{32}}\\\\\frac{5.65}{2}=\sqrt{\frac{M_1}{32}}\\\\M_1=255.38amu$

The gas having this molar mass is $C_{13}H_{21}NO_2S$