Question

for effusion of equal volume a gas takes twice the time as taken by methane under similar condition . The molar mass of the gas is

Answer 1

This is the answer...

Answer 2

The molar mass of the gas is 64 g/mol

Explanation:

From Graham's law of effusion rates, the rate of effusion of a gas is inversely proportional to the square root of its molar mass.

$\frac{rate_1}{rate_2}=\sqrt{\frac{M_2}{M_1}}$

$rate=\frac{volume}{time}$

Rate is volume effused per unit time. Since, the volumes are same, the formula could be written as:

$\frac{t_2}{t_1}=\sqrt{\frac{M_2}{M_1}}$

let's say in formula, subscript 1 is for unknown gas and 2 is for the methane gas.

Putting the values in the formula:

$\frac{t}{2t}=\sqrt{\frac{16}{M_1}}$

Squaring both sides:

$M_1=64g/mol$

The molar mass of the gas is 64 g/mol

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