Question

If some moles of $O_2$ diffuse in $18$ sec and same moles of an unknown gas diffuse in $45$ sec then what is the molecular weight of the unknown gas?

• A. $\frac{{45}^2}{{18}^2}\times 32$
• B. $\frac{{18}^2}{{45}^2}\times 16$
• C. $\frac{{18}^2}{{45}^2\times 32}$
• D. $\frac{{45}^2}{{18}^2\times 32}$

Answer 1

Answer: (A)

$\frac{{45}^2}{{18}^2}\times 32$

Given that

Time taken for diffusion of $O_{2}gas,t_{O_2}=$ 18 sec

Time taken for diffusion of unknown gas $X,t_X=$ 45 sec

Moles of $O_2$ gas diffused, $n_{O_2}=$ moles of unkown gas $X$ diffused, $n_X=n$

From Graham's law of diffusion we know --

$Rateofdiffusion\propto \frac{1}{\sqrt{M}}$ ----- (1)

Rate of diffusion, $r_N=\frac{\Delta n}{\Delta t}$ ------ (2)

Now, using (1) and (2) equation formula we have

$\frac{Rateofdiffusionof{O}_{2}gas}{RateofdiffusionofunknownXgas}=\frac{\frac{n_{O_2}}{t_{O_2}}}{\frac{n_X}{t_X}}=\frac{t_X}{t_{O_2}}=\sqrt{\frac{M_X}{M_{O_2}}}$ -------- (3)

where,

$M_X=$ Molar mass of unknown gas $X$

$M_{O_2}=$ Molar mass of $O_2$ gas = 32 g/mol

Substituting the values in (3) equation we have

$\frac{45sec}{18sec}=\sqrt{\frac{M_X}{32g/mol}}$

Squaring on both side we have

$\Rightarrow \frac{{45}^2}{{18}^2}=\frac{M_X}{32g/mol}$

$\Rightarrow \frac{{45}^2}{{18}^2}\times 32g/mol=M_X$

"or"

$\Rightarrow M_X=\frac{{45}^2}{{18}^2}\times 32g/mol$

$\therefore$ (A) option is correct.