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.
$$
\frac{t_X}{t_{O_2}} = \sqrt{\frac{M_X}{M_{O_2}}}
$$

$t_X = 28.3 \text{ s}$  
$t_{O_2} = 20 \text{ s}$  
$M_{O_2} = 2 \times 16 = 32 \text{ g/mol}$

$$
\frac{28.3}{20} = \sqrt{\frac{M_X}{32}}
$$
$$
1.415 = \sqrt{\frac{M_X}{32}}
$$

$$
(1.415)^2 = \frac{M_X}{32}
$$
$$
2.002225 = \frac{M_X}{32}
$$
$$
M_X = 2.002225 \times 32
$$
$$
M_X = 64.0712
$$

The molecular mass of gas X is $\boxed{\text{64.07 g/mol}}$.

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.

Distance for gas V ($d_V$) = $\frac{1}{5} \text{ m} = 0.2 \text{ m} = 20 \text{ cm}$  
Distance for gas W ($d_W$) = $10 \text{ cm}$

$t_V = 10 \text{ s}$  
$t_W = 10 \text{ s}$  
$M_V = 16.0 \text{ g/mol}$

$$
\frac{20}{10} = \sqrt{\frac{M_W}{16.0}}
$$
$$
2 = \sqrt{\frac{M_W}{16.0}}
$$

$$
2^2 = \frac{M_W}{16.0}
$$
$$
4 = \frac{M_W}{16.0}
$$
$$
M_W = 4 \times 16.0
$$
$$
M_W = 64.0
$$

The molecular mass of W is $\boxed{\text{64.0 g/mol}}$.

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.

$V_1 = 3.5 \times 10^{-2} \text{ m}^3$  
$T_1 = 291 \text{ K}$  
$P_1 = 1.0 \times 10^5 \text{ Pa}$  
$V_2 = 2.8 \times 10^{-2} \text{ m}^3$  
$P_2 = 1.0 \times 10^5 \text{ Pa}$ (pressure is constant)

$$
\frac{V_1}{T_1} = \frac{V_2}{T_2}
$$

$$
\frac{3.5 \times 10^{-2}}{291} = \frac{2.8 \times 10^{-2}}{T_2}
$$

$$
T_2 = \frac{(2.8 \times 10^{-2}) \times 291}{3.5 \times 10^{-2}}
$$
$$
T_2 = \frac{2.8 \times 291}{3.5}
$$
$$
T_2 = \frac{814.8}{3.5}
$$
$$
T_2 = 232.8
$$

The temperature at which the volume would be $2.8 \times 10^{-2} \text{ m}^3$ is $\boxed{\text{232.8 K}}$.

$M_{O_2} = 2 \times 16.0 = 32.0 \text{ g/mol}$  
$M_{SO_2} = 32.0 + (2 \times 16.0) = 32.0 + 32.0 = 64.0 \text{ g/mol}$

$V_{O_2} = 60 \text{ cm}^3$  
$t_{O_2} = 50 \text{ s}$  
$V_{SO_2} = 60