Question

A sample of an unknown gas effuses in 12.5 min. An equal volume of H2 in the same apparatus under the same conditions effuses in 2.42 min. What is the molar mass of the unknown gas

Answer 1

Answer:

53.4 gMol-1

Explanation:

Let the mass of the unknown gas be M

Let the molar mass of hydrogen gas be 2×1=2gMol-1

Time for diffusion of unknown gas = 12.5 min

Time for diffusion of hydrogen= 2.42 min

From Graham's law:

t1/t2=√M1/M2

Hence:

2.42/12.5= √2/M

Hence M= 53.4 gMol-1

Answer 2

Answer:

The molar mass of the unknown gas is 40.06 g/mol

Explanation:

Graham's law of effusion states that the rate of effusion of a gaseous substance is inversely proportional to the square root of its molar mass.

$\frac{R_{b} }{R_{a} } = \sqrt{\frac{M_{a} }{M_{b} } }$  $= \frac{t_{a} }{t_{b} }$ where R = rate of effusion, M = molar mass and t= time of effusion

⇒ $\sqrt{\frac{2 g/mol}{x g/mol} } = \frac{162secs}{725secs}$

x g/mol = $\frac{2}{0.223448^{2} }$

= 40.06 g/mol