Question

Some moles of oxygen diffused through a small opening in 18 s. Same number of moles of an unknown gas diffuses through the same opening in \(45 \mathrm{~s}\). Molccular weight of the unknown gas is (1) \((32)^{2} \times \frac{18}{45}\) (2) \((32)^{2} \times \frac{18}{(45)^{2}}\) (3) \((32) \times \frac{(45)^{2}}{(18)^{2}}\) (4) \((32)^{2} \times \frac{45}{18}\)

Short answer

The molecular weight of the unknown gas is 200 g/mol (option 3).

Step-by-step solution

- Understand Graham's Law

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. Mathematically, it is represented as: \ \[ \frac{r_{1}}{r_{2}} = \sqrt{\frac{M_{2}}{M_{1}}} \] where \( r_1 \) and \( r_2 \) are the rates of effusion of gases 1 and 2, and \( M_1 \) and \( M_2 \) are the molar masses of gases 1 and 2, respectively.

- Relate Rates with Time

The time taken for effusion is inversely proportional to the rate of effusion. Therefore, you can write: \ \[ \frac{r_{1}}{r_{2}} = \frac{t_{2}}{t_{1}} \] This means the ratio of the times taken for the gases to effuse is equal to the inverse of the ratio of their rates of effusion.

- Insert Known Values

Given that oxygen takes 18 s and the unknown gas takes 45 s, and that the molar mass of oxygen is 32 g/mol, substitute these values into the equation: \ \[ \frac{18}{45} = \sqrt{\frac{M_{unknown}}{32}} \]

- Isolate Molar Mass of Unknown Gas

Square both sides of the equation to remove the square root: \ \[ \left(\frac{18}{45}\right)^2 = \frac{M_{unknown}}{32} \] Solve for \( M_{unknown} \): \ \[ M_{unknown} = 32 \times \left(\frac{45}{18}\right)^2 \]

- Simplify the Expression

Simplify the fraction within the expression: \ \[ \frac{45}{18} = \frac{5}{2} \] Substitute this back into the equation: \ \[ M_{unknown} = 32 \times \left(\frac{5}{2}\right)^2 \] \ \[ M_{unknown} = 32 \times \frac{25}{4} \] \ \[ M_{unknown} = 32 \times 6.25 \] \ \[ M_{unknown} = 200 \]

Key concepts

Molecular Weight Calculation

Molecular weight, also known as molar mass, is a key concept in chemistry. It represents the mass of one mole of a particular substance. For gases, the molecular weight helps determine how many grams of the gas are in one mole. Knowing the molecular weight of a gas allows chemists to predict its behavior in reactions and various conditions. In the given problem, we need to find the molecular weight of an unknown gas by using Graham's Law of Effusion. The molecular weight of a known gas like oxygen serves as a reference to find the corresponding value for the unknown gas.

Rate of Effusion

Graham's Law of Effusion explains how gases pass through a small opening. The rate of effusion (how fast the gas escapes) depends inversely on the square root of its molecular weight. This means lighter gases effuse quicker than heavier ones. The mathematical representation is:
\[ \frac{r_{1}}{r_{2}} = \sqrt{\frac{M_{2}}{M_{1}}} \]
Here, \(r_1\) and \(r_2\) are the effusion rates of gases 1 and 2, and \(M_1\) and \(M_2\) are their molar masses. In our example, the rates of effusion can be linked to the time it takes each gas to effuse because time is inversely proportional to the rate of effusion. Therefore,
\[ \frac{r_{1}}{r_{2}} = \frac{t_{2}}{t_{1}} \]
By substituting the time values (18 s for oxygen and 45 s for the unknown gas), we can equate the ratios to eventually find the molecular weight of the unknown gas.

Proportional Relationships

Understanding proportional relationships is essential to solve the given problem. Graham's Law tells us that the rate of effusion (\(r\)) and the square root of the molecular weight (\(\sqrt{M}\)) are inversely proportional. When analyzing the given times for effusion, we applied this inversely proportional relationship.

• Step 3 shows the substitution: \[ \frac{18}{45} = \sqrt{\frac{M_{unknown}}{32}} \]
• Step 4 involves squaring both sides to isolate the molecular weight: \[ \left(\frac{18}{45}\right)^2 = \frac{M_{unknown}}{32} \]
• After simplifying, we find: \[ M_{unknown} = 32 \times \left(\frac{5}{2}\right)^2 = 200 \]

This approach highlights how proportional relationships allow us to switch between time, rate of effusion, and molecular weight to find unknown values.