At \(25^{\circ} \mathrm{C}\) and \(380 \mathrm{~mm} \mathrm{Hg}\), the density of sulfur dioxide is \(1.31 \mathrm{~g} / \mathrm{L}\). The rate of effusion of sulfur dioxide through an orifice is \(4.48 \mathrm{~mL} / \mathrm{s}\). What is the density of a sample of gas that effuses through an identical orifice at the rate of \(6.78 \mathrm{~mL} / \mathrm{s}\) under the same conditions? What is the molar mass of the gas?

Graham's Law of Effusion states that the ratio of effusion rates of two different gases is equal to the inverse ratio of the square roots of their molar masses: $$\frac{Rate_1}{Rate_2} = \frac{\sqrt{M_2}}{\sqrt{M_1}}$$ In this case, \(Rate_1\) and \(Rate_2\) are the effusion rates of sulfur dioxide and the unknown gas, respectively, and \(M_1\) and \(M_2\) are their molar masses.

Given the effusion rates for sulfur dioxide (4.48 mL/s) and the unknown gas (6.78 mL/s), calculate their ratio: $$\frac{Rate_1}{Rate_2} = \frac{4.48}{6.78} = 0.661 \mathrm{(rounded~to~3~decimal~places)}$$

In order to find the molar mass of the unknown gas, we need the molar mass of sulfur dioxide (SO2). To calculate it, add the molar masses of each atom found in SO2: $$M_1 = 1 \times M_S + 2 \times M_{O} = 32.07 + 2(16.00) = 64.07$$ where \(M_S = 32.07 g/mol\) (molar mass of sulfur) and \(M_O = 16.00 g/mol\) (molar mass of oxygen).

Using the ratio of effusion rates and the molar mass of sulfur dioxide, we can now solve for the molar mass of the unknown gas: $$\frac{Rate_1}{Rate_2} = \frac{\sqrt{M_2}}{\sqrt{M_1}}$$ $$M_2 = M_1 \left(\frac{Rate_2}{Rate_1}\right)^2 = 64.07 \times \left(\frac{6.78}{4.48}\right)^2 \approx 139.1 \mathrm{g/mol}$$

We know the molar mass and the effusion rate of the gas; in order to find its density, we first need to determine its molar volume. Since we are given the conditions of \(25^{\circ} \mathrm{C}\) and \(380 \mathrm{~mm} \mathrm{Hg}\), use the Ideal Gas Law (\(PV = nRT\)) to calculate the molar volume, where \(V\) is the volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is the temperature. First, convert the temperature and pressure to Kelvin and atmospheric pressure, respectively: $$T = 25^{\circ} \mathrm{C} + 273 = 298 \mathrm{K}$$ $$P = \frac{380 \mathrm{~mm} \mathrm{Hg}}{760 \mathrm{~mm} \mathrm{Hg/atm}} \approx 0.5 \mathrm{~atm}$$ Now, solve for the molar volume (\(V_{molar}\)) using the Ideal Gas Law: $$PV_{molar} = RT$$ $$V_{molar} = \frac{RT}{P} = \frac{0.0821 \mathrm{~L} \cdot \mathrm{atm/mol} \cdot \mathrm{K} \cdot 298 \mathrm{~K}}{0.5 \mathrm{~atm}} \approx 49.2 \mathrm{~L/mol}$$

Finally, calculate the density of the unknown gas by dividing its molar mass by its molar volume: $$Density = \frac{M_2}{V_{molar}} = \frac{139.1 \mathrm{g/mol}}{49.2 \mathrm{~L/mol}} \approx 2.83 \mathrm{g/L}$$ In conclusion, the density of the unknown gas is approximately 2.83 g/L, and its molar mass is approximately 139.1 g/mol.