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Difficulty: HardGraham's Law of Diffusion and Effusion

A 160 cm3160\text{ cm}^3 sample of an unknown gas ZZ diffuses through a porous partition in 30 seconds30\text{ seconds}. Under identical conditions of temperature and pressure, an 80 cm380\text{ cm}^3 sample of sulfur(IV) oxide (SO2SO_2) gas diffuses through the same partition in 20 seconds20\text{ seconds}. What is the relative molecular mass of gas ZZ? [Relative atomic masses: S=32S = 32, O=16O = 16]

  1. A
    16 g/mol16\text{ g/mol}
  2. 36 g/mol36\text{ g/mol}Answer
  3. C
    48 g/mol48\text{ g/mol}
  4. D
    85.3 g/mol85.3\text{ g/mol}

Answer

The relative molecular mass of gas Z is 36 g/mol36\text{ g/mol}.
The rate of diffusion for gas Z is 160/30=16/3 cm3/s160/30 = 16/3\text{ cm}^3/\text{s} and for SO2SO_2 is 80/20=4 cm3/s80/20 = 4\text{ cm}^3/\text{s}. According to Graham's law, (rZ/rSO2)2=MSO2/MZ(r_Z / r_{SO_2})^2 = M_{SO_2} / M_Z. Substituting the values gives (4/3)2=16/9=64/MZ(4/3)^2 = 16/9 = 64 / M_Z, yielding MZ=36 g/molM_Z = 36\text{ g/mol}.

Step-by-Step Solution

1
Calculate the molar mass of sulfur(IV) oxide (SO2SO_2)
MSO2=32+(2×16)=64 g/molM_{SO_2} = 32 + (2 \times 16) = 64\text{ g/mol}
Molar mass of the reference gas is required for Graham's Law calculation.
2
Determine the rates of diffusion for gas ZZ and SO2SO_2
rZ=160 cm330 s=163 cm3/sr_Z = \frac{160\text{ cm}^3}{30\text{ s}} = \frac{16}{3}\text{ cm}^3/\text{s} and rSO2=80 cm320 s=4 cm3/sr_{SO_2} = \frac{80\text{ cm}^3}{20\text{ s}} = 4\text{ cm}^3/\text{s}
Rate of diffusion is defined as volume of gas diffused per unit time (r=V/tr = V/t).
3
Calculate the ratio of the diffusion rates
rZrSO2=16/34=43\frac{r_Z}{r_{SO_2}} = \frac{16/3}{4} = \frac{4}{3}
Comparing the two rates simplifies substitution into Graham's Law equation.
4
Apply Graham's Law of Diffusion to solve for the unknown molar mass (MZM_Z)
\frac{r_Z}{r_{SO_2}} = \sqrt{\frac{M_{SO_2}}{M_Z}} \implies \left(\frac{4}{3}\right)^2 = \frac{64}{M_Z} \implies \frac{16}{9} = \frac{64}{M_Z} \implies M_Z = \frac{64 \times 9}{16} = 36\text{ g/mol}
According to Graham's Law, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass.

Key Concept

Graham's Law of Diffusion relates gas diffusion rates to their molar masses: r1/r2=M2/M1r_1 / r_2 = \sqrt{M_2 / M_1}.
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Graham's Law of Diffusion and Effusion

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