Question

The rate of effusion of an unknown gas was measured and found to be 13.9 mL/min. Under identical conditions, the rate of effusion of pure oxygen (O₂)gas is 16.3 mL/min. Based on this information, the identity of the unknown gas could be:

O F₂
O NO
O N₂OC₂H₆
O None of the other given answers are correct.

Answer 1

According to Graham's law of effusion, the molar mass of the unknown gas is approximately 27.29 g/mol, which suggests that it could be NO (nitric oxide).

option b is th correct

Step-by-step explanation:

According to Graham's law of effusion, the rate of effusion of a gas is inversely proportional to the square root of its molar mass. In this case, the rate of effusion of the unknown gas is 13.9 mL/min and the rate of effusion of pure oxygen is 16.3 mL/min. To determine the molar mass of the unknown gas, we can set up the following equation:

(rate of effusion unknown gas) / (rate of effusion oxygen) = sqrt(molar mass oxygen) / sqrt(molar mass unknown gas)

Plugging in the given values, we have:

(13.9 mL/min) / (16.3 mL/min) = sqrt(32 g/mol) / sqrt(molar mass unknown gas)

Simplifying and solving for the unknown gas molar mass, we get:

molar mass unknown gas = (32 g/mol) * (13.9 mL/min) / (16.3 mL/min)

molar mass unknown gas = 27.29 g/mol

Based on this molar mass, the identity of the unknown gas could be NO (nitric oxide), which has a molar mass of approximately 30 g/mol.