Question

A sample of nitrogen gas is contaminated with a gas (gas A) of unknown molar mass. The partial pressure of each gas is known to be 200 torr at 25°C. The gases are allowed to effuse through a pinhole, and it is found that gas A escapes 2.65 times the rate of nitrogen gas. What is the molar mass of gas A?

Answer 1

The molar mass of gas A, which effuses 2.65 times faster than nitrogen, is calculated using Graham's law of effusion and is found to be 3.99 g/mol.

Step-by-step explanation:

The question involves using Graham's law of effusion to determine the molar mass of gas A, which effuses 2.65 times faster than nitrogen gas. Graham's law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass (rate √M = constant). Since the rate of gas A is 2.65 times the rate of nitrogen (N₂), we can set up the following equation derived from Graham's law:

(rate A / rate N₂) = (√(molar mass N₂) / √(molar mass A))

Given that the molar mass of N₂ is 28.01 g/mol, we can rearrange the equation and solve for the molar mass of gas A:

(2.65) = (√(28.01) / √(molar mass A))

(2.65)² = (28.01 / molar mass A)

molar mass A = (28.01 / (2.65)²)

molar mass A = 28.01 / 7.0225

molar mass A = 3.99 g/mol