Question

an unknown gas effuses at a rate 2.00 times the rate of so₃. what is the molar mass of the unknown gas?

Answer 1

Answer: 20.0 g/mol

Explanation:

$\frac{2.00}{1.00} = \frac{\sqrt{80.06 g/mol} }{\sqrt{mm_{A}} }$

$mm_{A} = \frac{80.06 g/mol}{2^{2} }$

$mm_{A} = 20.015 g/mol$

Rounded to sig figs: 20.0 g/mol

Answer 2

Using the Graham's law of diffusion, the molar mass of the unknown gas is 20 g/mol.

Graham's law

Graham's law gives the relationship between the molar mass of a gas and the rate of effusion of the gas.

Let Rx be the rate of effusion of the unknown gas

RSO3 be the rate of effusion of SO3

Mx be the molar mass of the unknown gas

MSO3 be the molar mass of SO3

Then;

RX/RSO3 = √MSO3/MX

2/1 = √80/Mx

4/1 = 80/Mx

Mx = 20

The molar mass of the unknown gas is 20 g/mol

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