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?
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 --
$$ Rate\ of\ diffusion \propto \dfrac{1}{\sqrt{M}} $$ ----- (1)
Rate of diffusion, $$r_N = \dfrac{\Delta n} {\Delta t}$$ ------ (2)
Now, using (1) and (2) equation formula we have
$$\dfrac{Rate \ of \ diffusion \ of \ O_2 \ gas}{Rate \ of \ diffusion \ of \ unknown \ X \ gas} = \dfrac{\dfrac{n_{O_2}}{t_{O_2}}}{\dfrac{n_X}{t_X}} = \dfrac{t_X}{t_{O_2}} = \sqrt{\dfrac{M_X}{M_{O_2}}}$$ -------- (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
$$\dfrac{45\ sec}{18\ sec} = \sqrt{\dfrac{M_X}{32\ g/mol}}$$
Squaring on both side we have
$$\Rightarrow \dfrac{45^2}{18^2} = \dfrac{M_X}{32\ g/mol}$$
$$\Rightarrow \dfrac{45^2}{18^2} \times 32\ g/mol = M_X$$
"or"
$$\Rightarrow M_X = \dfrac{45^2}{18^2} \times 32\ g/mol$$
$$\therefore$$ (A) option is correct.
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