Question

A gas diffuses 1/6 times faster than hydrogen gas (H2). What is the molar mass of the gas? 59.95 g/mol 66.54 g/mol 68.68 g/mol 72.58 g/mol

Answer 1

Solution us here,

let the rate of diffusion of H2(r1) be x and⁷ the rate of diffusion of gas(r2) be x+x/6=7x/6.

Molar mass of H2 (M1)=2 g

Molar mass of gas(M2)= ?

Now, from the Ghram's law of diffusion of gas,

$\frac{r1}{r2} = \sqrt{ \frac{m2}{m1} } \\ or \: \frac{x}{ \frac{7x}{6} } = \sqrt{ \frac{m2}{2} } \\ or \: \frac{6}{7} = \sqrt{ \frac{m2}{2} } \\ squaring \: in \: both \: sides \\ \frac{36}{49} = \frac{m2}{2} \\ or m2 = 1 .469$

here, it looks like I have done wrong.

But all of the answers in option are wrong.

Because, rate of diffusion of any gas is inversely proportional to the square root of its malor mass.

And in the question, gas has higher rate of diffusion than hrdrogen. So it should have molar mass less than hydrogen.

Answer 2

Answer: The molar mass of unknown gas is 72.58 g/mol

Explanation:

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows:

$\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}$

Molar mass of hydrogen= 2.02 g/mol

Molar mass of unknown gas= X g/mol

For the rate of diffusion of hydrogen to unknown gas, we write the expression:

$\frac{Rate_{X}}{Rate_{H_2}}=\sqrt{\frac{M_{H_2}}{M_{X}}}$

$\frac{1}{6}=\sqrt{\frac{2.02}{M_{X}}$

$M_{X}=72.58g/mol$

Hence, the molar mass of unknown gas is 72.58 g/mol