Question

A (1.820x10⁻⁵) mol sample of "gas a," which has a molar mass of (3.14x10¹) g/mol takes (2.7330x10²) s to effuse through a pinhole. a sample of (1.820x10⁻⁵) mol of "gas b" takes (1.838x10²) s to effuse though the same pinhole under the same temperature and pressure conditions as "gas a". what is the molar mass of "gas b"? note: your answer is assumed to be reduced to the highest power possible.

Answer 1

To find the molar mass of Gas B, we can use Graham's law of effusion and set up a ratio using the rates of effusion and molar masses of the two gases. Solving for the molar mass of Gas B, we find the answer to be approximately X g/mol.

Step-by-step explanation:

Graham's law of effusion states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. In this case, we are given that Gas A, with a molar mass of 3.14x10¹ g/mol, takes 2.7330x10² s to effuse through a pinhole, while Gas B takes 1.838x10² s to effuse under the same conditions. To find the molar mass of Gas B, we can set up a ratio using Graham's law:

(Rate of A / Rate of B) = sqrt((Molar Mass of B / Molar Mass of A))

Substituting the given values, we get:

(2.7330x10² / 1.838x10²) = sqrt((Molar Mass of B / 3.14x10¹))

Simplifying and solving for the molar mass of B, we find:

Molar Mass of B = (2.7330x10² / 1.838x10²) * sqrt(3.14x10¹)

Rounding to the highest power possible, the molar mass of Gas B is approximately X g/mol.