Question

A gas effuses 1.55 times faster than propane (C3H8)at the

Answer 1

Answer:  The mass of the gas is 18.3 g/mol.

Explanation:

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

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

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

$\frac{Rate_{X}}{Rate_{C_3H_8}}=1.55$

$\frac{Rate_{X}}{Rate_{C_3H_8}}=\sqrt{\frac{M_{C_3H_8}}{M_{X}}}$

$1.55=\sqrt{\frac{44}{M_{X}}$

Squaring both sides and solving for $M_{X}$

$M_{X}=18.3g/mol$

Hence, the molar mas of unknown gas is 18.3 g/mol.

Answer 2

Answer: The molar mass of the unknown gas is 18.3 g/mol

Explanation:

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

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

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

We are given:

$\text{Rate}_{\text{(unknown gas)}}=1.55\times \text{Rate}_{C_3H_8}$

We know that:

Molar mass of propane = 44 g/mol

Taking the ratio of the rate of effusion of the gases, we get:

$\frac{\text{Rate}_{\text{(unknwon gas)}}}{\text{Rate}_{C_3H_8}}=\sqrt{\frac{M_{C_3H_8}}{M_{\text{(unknown gas)}}}}$

Putting values in above equation, we get:

$\frac{1.55\times \text{Rate}_{C_3H_8}}{\text{Rate}_{C_3H_8}}=\sqrt{\frac{44}{M_{\text{(unknown gas)}}}}$

$1.55=\sqrt{\frac{44}{M_{\text{(unknown gas)}}}}\\\\1.55^2=\frac{44}{M_{\text{unknwon gas}}}\\\\M_{\text{unknwon gas}}=\frac{44}{2.4025}\\\\M_{\text{unknwon gas}}=18.3g/mol$

Hence, the molar mass of the unknown gas is 18.3 g/mol