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Chapter 5

Q. 5.11

In an effusion experiment, argon gas is allowed to expand through a tiny opening into an evacuated flask of volume 120 mL for 32.0 s, at which point the pressure in the flask is found to be 12.5 mm Hg. This experiment is repeated with a gas X of unknown molar mass at the same T and P. It is found that the pressure in the flask builds up to 12.5 mm Hg after 48.0 s. Calculate the molar mass of X.

ANALAYSIS
volume of both flasks (120 mL); pressure in both flasks (12.5 mm Hg);
time for Ar effusion (32.0 s); time for gas (X) effusion (48.0 s)
Information given:
Temperature, pressure, and volume are the same for both flasks.
rate of effusion for each gas
MM of argon
Information implied:
MM of X Asked for:

STRATEGY

1. Since T, P, and V are the same for both gases, the number of moles of gas in both flasks is the same.
n_{Ar}  =   n_{X} = n
2. The rate of effusion is in mol/time.

rate = \frac{n_{Ar}}{time} = \frac{n_{X}}{time} = \frac{n}{time}

3. Substitute into Graham’s law of effusion, where A = gas X and B = Ar.

\frac{rate  B}{rate  A} = \left(\frac{MM_{A}}{MM_{B}} \right)^{1/2}   →  \frac{rate  Ar}{rate  X} = \left(\frac{MM_{X}}{MM_{Ar}} \right)^{1/2}

Step-by-Step

Verified Solution

rate X = \frac{n}{48.0  s}              rate Ar = \frac{n}{32.0  s} rates
\frac{\frac{n}{32.0  s}}{\frac{n}{48.0  s}} = \left(\frac{MM_{X}}{39.95  g/mol} \right)^{1/2}   →   1.50 = \left(\frac{MM_{X}}{39.95  g/mol} \right)^{1/2}

(1.50)² = \left(\left(\frac{MM_{X}}{39.95  g/mol} \right)^{1/2}\right) ^{2}    →   2.25 = \frac{MM_{X}}{39.95  g/mol}  →   MM_{X} = 89.9 g/mol

MM_{X}

END POINT

Since the unknown gas takes longer to effuse, it should have a larger molar mass than argon. It does!