A "canned heat" product used to warm buffet dishes consists of a homogeneous mixture of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) and paraffin, which has an average formula of \(\mathrm{C}_{24} \mathrm{H}_{50}\). What mass of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) should be added to \(620 \mathrm{~kg}\) of the paraffin to produce 8 torr of ethanol vapor pressure at \(35^{\circ} \mathrm{C}\) ? The vapor pressure of pure ethanol at \(35^{\circ} \mathrm{C}\) is 100 torr.

We know: - mass of paraffin = 620 kg - vapor pressure of ethanol = 8 torr - vapor pressure of pure ethanol at 35°C = 100 torr Using Raoult's Law: \[P_{ethanol} = x_{ethanol} \times P^{0}_{ethanol}\] where: - \(P_{ethanol}\) is the vapor pressure of ethanol in the solution - \(x_{ethanol}\) is the mole fraction of ethanol in the solution - \(P^{0}_{ethanol}\) is the vapor pressure of pure ethanol at 35°C We need to find \(x_{ethanol}\) and then determine the mass of ethanol required.

From Raoult's Law, we can rearrange the equation to find the mole fraction of ethanol: \[x_{ethanol} = \frac{P_{ethanol}}{P^{0}_{ethanol}}\] Plugging in the given values: \[x_{ethanol} = \frac{8 \ \text{torr}}{100 \ \text{torr}}\] \[x_{ethanol} = 0.08\]

We know the mole fraction of ethanol in the solution is 0.08. Suppose we have \(n_{paraffin}\) moles of paraffin and \(n_{ethanol}\) moles of ethanol: \[x_{ethanol} = \frac{n_{ethanol}}{n_{paraffin} + n_{ethanol}}\] Rearrange the equation to solve for \(n_{ethanol}\): \[n_{ethanol} = x_{ethanol}(n_{paraffin} + n_{ethanol})\] We know the mass of paraffin is 620 kg, and the molecular weight of paraffin is 338.62 g/mol (\(24 \times 12.01 + 50 \times 1.01\)). Calculate the moles of paraffin: \[n_{paraffin} = \frac{620 \ \text{kg} \times 1000 \ \frac{\text{g}}{\text{kg}}}{338.62 \ \frac{\text{g}}{\text{mol}}}\] \[n_{paraffin} = 1830.87 \ \text{mol}\] Substitute this value and the mole fraction of ethanol (0.08) back into the equation: \[n_{ethanol} = 0.08(n_{paraffin} + n_{ethanol})\] \[n_{ethanol} = 0.08(1830.87 + n_{ethanol})\]

Solve for \(n_{ethanol}\) in the equation above: \[n_{ethanol} = \frac{0.08(1830.87)}{1 - 0.08}\] \[n_{ethanol} = 159.90 \ \text{mol}\]

Now that we have the moles of ethanol, we can calculate the mass of ethanol required. The molecular weight of ethanol is 46.07 g/mol (\(2 \times 12.01 + 6 \times 1.01 + 16\)). Use the formula: mass = moles × molecular weight \[m_{ethanol} = n_{ethanol} \times M_{ethanol}\] \[m_{ethanol} = 159.90 \ \text{mol} \times 46.07 \frac{\text{g}}{\text{mol}}\] \[m_{ethanol} = 7366 \ \text{g}\] Therefore, 7366 g (approximately 7.37 kg) of ethanol should be added to 620 kg of paraffin to produce 8 torr of ethanol vapor pressure at 35°C.