State Graham's Law of Diffusion.

You're on a roll — here are the solutions to your chemistry questions: 1. A sample of unknown compound gas X is shown by analysis to contain Sulphur and Oxygen. The gas requires 28.3 seconds to diffuse through a small aperture into a vacuum. An identical number of oxygen molecules pass through the same aperture in 20 seconds. Determine the molecular mass of gas X (O=16, S=32). Step 1: State Graham's Law of Diffusion. Graham's Law states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. $$ \frac{t_X}{t_{O_2}} = \sqrt{\frac{M_X}{M_{O_2}}} $$ Step 2: Identify the given values. $t_X = 28.3 \text{ s}$ $t_{O_2} = 20 \text{ s}$ $M_{O_2} = 2 \times 16 = 32 \text{ g/mol}$ Step 3: Substitute the values into the formula. $$ \frac{28.3}{20} = \sqrt{\frac{M_X}{32}} $$ $$ 1.415 = \sqrt{\frac{M_X}{32}} $$ Step 4: Square both sides to solve for $M_X$. $$ (1.415)^2 = \frac{M_X}{32} $$ $$ 2.002225 = \frac{M_X}{32} $$ $$ M_X = 2.002225 \times 32 $$ $$ M_X = 64.0712 $$ The molecular mass of gas X is $\boxed{\text{64.07 g/mol}}$. 2. a) State Graham's Law of diffusion. Graham's Law of diffusion states that the rate of diffusion or effusion of a gas is inversely proportional to the square root of its molar mass, assuming constant temperature and pressure. b) Gas V takes 10 seconds to diffuse through a distance of one fifth of a meter. Another gas W takes the same time to diffuse through a distance of 10 cm. If the relative molecular mass of gas V is 16.0; calculate the molecular mass of W. Step 1: Convert all distances to the same unit (cm). Distance for gas V ($d_V$) = $\frac{1}{5} \text{ m} = 0.2 \text{ m} = 20 \text{ cm}$ Distance for gas W ($d_W$) = $10 \text{ cm}$ Step 2: Identify the given values. $t_V = 10 \text{ s}$ $t_W = 10 \text{ s}$ $M_V = 16.0 \text{ g/mol}$ Step 3: Apply Graham's Law. Since time is constant, the rate of diffusion is proportional to the distance diffused. $$ \frac{r_V}{r_W} = \frac{d_V}{d_W} = \sqrt{\frac{M_W}{M_V}} $$ Step 4: Substitute the values into the formula. $$ \frac{20}{10} = \sqrt{\frac{M_W}{16.0}} $$ $$ 2 = \sqrt{\frac{M_W}{16.0}} $$ Step 5: Square both sides to solve for $M_W$. $$ 2^2 = \frac{M_W}{16.0} $$ $$ 4 = \frac{M_W}{16.0} $$ $$ M_W = 4 \times 16.0 $$ $$ M_W = 64.0 $$ The molecular mass of W is $\boxed{\text{64.0 g/mol}}$. 3. a) State Charles' Law. Charles' Law states that for a fixed mass of gas at constant pressure, the volume of the gas is directly proportional to its absolute temperature. b) The volume of a sample of nitrogen gas at a temperature of 291 K and $1.0 \times 10^5$ Pascals was $3.5 \times 10^{-2} \text{ m}^3$. Calculate the temperature at which the volume of the gas would be $2.8 \times 10^{-2} \text{ m}^3$ at $1.0 \times 10^5$ Pascals. Step 1: Identify the given values. $V_1 = 3.5 \times 10^{-2} \text{ m}^3$ $T_1 = 291 \text{ K}$ $P_1 = 1.0 \times 10^5 \text{ Pa}$ $V_2 = 2.8 \times 10^{-2} \text{ m}^3$ $P_2 = 1.0 \times 10^5 \text{ Pa}$ (pressure is constant) Step 2: Apply Charles' Law. $$ \frac{V_1}{T_1} = \frac{V_2}{T_2} $$ Step 3: Substitute the values into the formula. $$ \frac{3.5 \times 10^{-2}}{291} = \frac{2.8 \times 10^{-2}}{T_2} $$ Step 4: Solve for $T_2$. $$ T_2 = \frac{(2.8 \times 10^{-2}) \times 291}{3.5 \times 10^{-2}} $$ $$ T_2 = \frac{2.8 \times 291}{3.5} $$ $$ T_2 = \frac{814.8}{3.5} $$ $$ T_2 = 232.8 $$ The temperature at which the volume would be $2.8 \times 10^{-2} \text{ m}^3$ is $\boxed{\text{232.8 K}}$. 4. $60 \text{ cm}^3$ of oxygen gas diffused through a porous partition in 50 seconds. How long would it take $60 \text{ cm}^3$ of sulphur (IV) oxide gas to diffuse through the same partition under the same conditions? (S=32.0, O=16.0) Step 1: Calculate the molar masses. $M_{O_2} = 2 \times 16.0 = 32.0 \text{ g/mol}$ $M_{SO_2} = 32.0 + (2 \times 16.0) = 32.0 + 32.0 = 64.0 \text{ g/mol}$ Step 2: Identify the given values. $V_{O_2} = 60 \text{ cm}^3$ $t_{O_2} = 50 \text{ s}$ $V_{SO_2} = 60