Problem 50 - Chemistry Central Science

Question

(a) Calculate the density of sulfur hexafluoride gas at 707 torr and \(21^{\circ} \mathrm{C}\). (b) Calculate the molar mass of a vapor that has a density of \(7.135 \mathrm{~g} / \mathrm{L}\) at \(12{ }^{\circ} \mathrm{C}\) and 743 torr.

Step-by-Step Solution

Answer

(a) The density of sulfur hexafluoride gas at 707 torr and \(21^{\circ} \mathrm{C}\) is approximately \(4.00 \mathrm{~g/L}\). (b) The molar mass of the vapor with a density of \(7.135 \mathrm{~g} / \mathrm{L}\) at \(12{ }^{\circ} \mathrm{C}\) and 743 torr is approximately \(96.07 \mathrm{~g/mol}\).

Step 1: Write down the Ideal Gas Law formula

The Ideal Gas Law formula is \(PV=nRT\), where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

Step 2: Convert temperature to Kelvin

To convert the given temperature of \(21^{\circ} \mathrm{C}\) to Kelvin, use the formula: \(T(K) = T(^{\circ} \mathrm{C}) + 273.15\). Therefore, \(T = 21 + 273.15 = 294.15\mathrm{~K}\).

Step 3: Convert pressure to atm

To convert the given pressure of 707 torr to atm, use the conversion factor 1 atm = 760 torr. Therefore, \(P = \frac{707}{760} \mathrm{~atm}\).

Step 4: Rewrite the formula in terms of density

Density (\(\rho\)) is defined as \(\rho = \frac{m}{V}\), where m is the mass. We can also represent mass as the product of moles and molar mass (M): \(m = nM\). Substituting this expression in the density formula, we get: \(\rho = \frac{nM}{V}\). Now, rearrange the Ideal Gas Law formula to obtain \(\frac{n}{V} = \frac{P}{RT}\) and substitute this expression into the density formula: \(\rho = \frac{PM}{RT}\).

Step 5: Substitute values and calculate density

Use the sulfur hexafluoride molar mass, M = 146.06 g/mol, and the given temperature and pressure values, \(T = 294.15\mathrm{~K}\) and \(P = \frac{707}{760} \mathrm{~atm}\), in the density formula: \(\rho = \frac{PM}{RT}\). We have: \(\rho = \frac{(\frac{707}{760})(146.06)}{(0.0821)(294.15)}\). On calculating, we get \(\rho \approx 4.00 \mathrm{~g/L}\). (a) The density of sulfur hexafluoride gas at 707 torr and \(21^{\circ} \mathrm{C}\) is approximately \(4.00 \mathrm{~g/L}\). (b) Calculate the molar mass of a vapor that has a density of \(7.135 \mathrm{~g} / \mathrm{L}\) at \(12{ }^{\circ} \mathrm{C}\) and 743 torr.

Step 6: Convert temperature to Kelvin

To convert the given temperature of \(12^{\circ} \mathrm{C}\) to Kelvin, use the formula: \(T(K) = T(^{\circ} \mathrm{C}) + 273.15\). Therefore, \(T = 12 + 273.15 = 285.15\mathrm{~K}\).

Step 7: Convert pressure to atm

To convert the given pressure of 743 torr to atm, use the conversion factor 1 atm = 760 torr. Therefore, \(P = \frac{743}{760} \mathrm{~atm}\).

Step 8: Rearrange the formula to find molar mass

Use the density formula from Step 4: \(\rho = \frac{PM}{RT}\). Rearrange it to find molar mass, M: \(M = \frac{\rho RT}{P}\).

Step 9: Substitute values and calculate molar mass

Use the given density (\(\rho = 7.135 \mathrm{~g/L}\)), temperature (\(T = 285.15\mathrm{~K}\)), and pressure (\(P = \frac{743}{760} \mathrm{~atm}\)), in the molar mass formula: \(M = \frac{\rho RT}{P}\). We have: \(M = \frac{(7.135)(0.0821)(285.15)}{(\frac{743}{760})}\). On calculating, we get \(M \approx 96.07 \mathrm{~g/mol}\). (b) The molar mass of the vapor with a density of \(7.135 \mathrm{~g} / \mathrm{L}\) at \(12{ }^{\circ} \mathrm{C}\) and 743 torr is approximately \(96.07 \mathrm{~g/mol}\).