A given volume of nitrogen, N2, rewuired 68.3 s to effuse from a hole in a chamber. Under the same conditions, another gas required 85.6 s for the samw volume to effuse. What is the molecular mass ?

To find the molecular mass of the second gas, we can use Graham's law of effusion which states that the rate of effusion of a gas is inversely proportional to the square root of its molecular mass.

Let's denote the molecular mass of the first gas as M1 and the molecular mass of the second gas as M2. The rate of effusion for the first gas (R1) is given by:

R1 = 1/√M1

Similarly, the rate of effusion for the second gas (R2) is given by:

R2 = 1/√M2

Given that the first gas took 68.3 s to effuse and the second gas took 85.6 s, we can set up the following equation:

R1/R2 = √(M2/M1)

Substituting the values:

(1/√M1)/(1/√M2) = √(M2/M1)

Simplifying the equation:

√(M2/M1) = √(68.3/85.6)

Squaring both sides:

M2/M1 = (68.3/85.6)^2

M2/M1 = 0.683^2

M2/M1 = 0.467

Now, we can rearrange the equation to solve for M2:

M2 = 0.467 * M1

Since we don't know the molecular mass of the first gas, we can choose an arbitrary value of M1. Let's assume M1 = 28 (molecular mass of nitrogen).

M2 = 0.467 * 28

M2 ≈ 13.076

Therefore, the molecular mass of the second gas is approximately 13.076.

To determine the molecular mass of the other gas, we can use Graham's law of effusion. Graham's law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.

In this case, we have two gases, nitrogen (N2) and the other gas. Let's call the molecular mass of the other gas M.

According to Graham's law, the rate of effusion of nitrogen (R_nitrogen) is inversely proportional to the square root of its molar mass (M_nitrogen):

R_nitrogen = k / √M_nitrogen

Similarly, the rate of effusion of the other gas (R_other) is inversely proportional to the square root of its molar mass (M_other):

R_other = k / √M_other

Since the given volume, temperature, and pressure are the same for both gases, the ratio of their rates of effusion (R_other/R_nitrogen) can be expressed as:

R_other / R_nitrogen = √(M_nitrogen / M_other)

Substituting the given times (68.3 s and 85.6 s) into the formula, we get:

√(M_nitrogen / M_other) = 85.6 / 68.3

Squaring both sides of the equation, we have:

M_nitrogen / M_other = (85.6 / 68.3)^2

Now, solving for M_other, the molecular mass of the other gas:

M_other = M_nitrogen / (85.6 / 68.3)^2

The molecular mass of nitrogen (M_nitrogen) is 28.013 g/mol.

Plugging in this value and calculating, we find:

M_other = 28.013 / (85.6 / 68.3)^2

Calculating this expression gives us the molecular mass of the other gas.

(rate1/rate2) = sqrt(M2/M1)

Substitute and solve for molar mass.

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