Question

State and explain Graham's law of diffusion. 120 \(\mathrm{cc}\) of \(\mathrm{SO}_{2}\) gas is diffused through a porous membrane in \(40 \mathrm{~min}\). Under similar conditions, \(360 \mathrm{cc}\) of another gas is diffused in \(30 \mathrm{~min}\). What is the molecular weight of the other gas?

Short answer

Question: Given that 120 cc of SO₂ gas diffuses in 40 minutes, and 360 cc of an unknown gas diffuses in 30 minutes, use Graham's law of diffusion to find the molecular weight of the unknown gas. Answer: The molecular weight of the unknown gas is 4 g/mol.

Step-by-step solution

Recall Graham's Law of Diffusion

Graham's law of diffusion states that the rate of diffusion of a gas is inversely proportional to the square root of its molecular weight. Mathematically, it can be expressed as: $$\frac{rate_{1}}{rate_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}$$, where \(rate_{1}\) and \(rate_{2}\) are the rates of diffusion for gases 1 and 2, respectively, and \(M_{1}\) and \(M_{2}\) are their respective molecular weights.

Calculate the rates of diffusion for both gases

We are given that 120 cc of SO\(_{2}\) gas diffuses in 40 minutes, while 360 cc of the unknown gas diffuses in 30 minutes. To find the rates of diffusion, we need to divide the volume by the time for both gases. $$rate_{SO_2} = \frac{120 \mathrm{cc}}{40 \mathrm{min}} = 3 \frac{\mathrm{cc}}{\mathrm{min}}$$ $$rate_{unknown} = \frac{360 \mathrm{cc}}{30 \mathrm{min}} = 12 \frac{\mathrm{cc}}{\mathrm{min}}$$

Apply Graham's Law of diffusion

Now we have the rates of diffusion for both gases. We use the formula for Graham's Law to create a proportion to find the molecular weight of the unknown gas, where \(M_{SO_2} = 64 \mathrm{g/mol}\) and \(rate_{SO_2}\) and \(rate_{unknown}\) are the rates calculated above. $$\frac{3}{12} = \sqrt{\frac{M_{unknown}}{64}}$$

Solve for the molecular weight of the unknown gas

We can now solve for the molecular weight of the unknown gas: $$\frac{1}{4} = \sqrt{\frac{M_{unknown}}{64}}$$ Square both sides of the equation: $$\frac{1}{16} = \frac{M_{unknown}}{64}$$ Now, multiply both sides by 64: $$64 \times \frac{1}{16} = M_{unknown}$$ Simplify the expression: $$M_{unknown} = 4$$ The molecular weight of the unknown gas is 4 g/mol.