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 $1.07 \mathrm{kPa}\( of ethanol vapor pressure at \)35^{\circ} \mathrm{C}$ ? The vapor pressure of pure ethanol at \(35^{\circ} \mathrm{C}\) is \(13.3 \mathrm{kPa}\).

Raoult's law states that the partial pressure of a component in a mixture is equal to the mole fraction of that component multiplied by the vapor pressure of the pure component. Mathematically, it can be represented as: \( P_A = x_A P_A^* \) where, \(P_A\) is the partial pressure of component A in the mixture, \(x_A\) is the mole fraction of component A in the mixture, and \(P_A^*\) is the vapor pressure of pure component A.

In the given problem, the vapor pressure of ethanol in the mixture is 1.07 kPa, and the vapor pressure of pure ethanol is 13.3 kPa. We can use Raoult's law to determine the mole fraction of ethanol in the mixture: \(1.07 \mathrm{kPa} = x_{\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}} \times 13.3 \mathrm{kPa} \) Solving for \( x_{\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}} \): \( x_{\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}} = 0.0805 \)

To find the mass of ethanol that should be added to the mixture, we need to determine the moles of paraffin and ethanol. The mass of paraffin is given as 620 kg, and the average molecular formula of paraffin is C24H50. First, let's find the molar mass of paraffin: \( \mathrm{MM}_{\mathrm{C}_{24}\mathrm{H}_{50}} = 24 \times 12.01 + 50 \times 1.01 = 338.34 \hspace{2mm} \mathrm{g/mol} \) Now, we can find the moles of paraffin: \(\mathrm{moles}_{\mathrm{C}_{24}\mathrm{H}_{50}} = \frac{620,000 \hspace{2mm} \mathrm{g}}{338.34 \hspace{2mm} \mathrm{g/mol}} = 1833.49 \hspace{2mm} \mathrm{mol} \) Now, let's find the moles of ethanol using the mole fraction of ethanol: \( \frac{\mathrm{moles}_{\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}}}{\mathrm{moles}_{\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}} + \mathrm{moles}_{\mathrm{C}_{24}\mathrm{H}_{50}}} = 0.0805 \) Solving for the moles of ethanol: \( \mathrm{moles}_{\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}} = 0.0805 \times \mathrm{moles}_{\mathrm{C}_{24}\mathrm{H}_{50}} / (1 - 0.0805) = 160.41 \hspace{2mm} \mathrm{mol} \)

Finally, we can find the mass of ethanol to be added by using the molar mass of ethanol, which is: \( \mathrm{MM}_{\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}} = 2 \times 12.01 + 6 \times 1.01 + 1 \times 16 = 46.07 \hspace{2mm} \mathrm{g/mol} \) The mass of ethanol to be added is: \( \mathrm{mass}_{\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}} = \mathrm{moles}_{\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}} \times \mathrm{MM}_{\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}} \) \( \mathrm{mass}_{\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}} = 160.41 \hspace{2mm} \mathrm{mol} \times 46.07 \hspace{2mm} \mathrm{g/mol} = 7,389.02 \hspace{2mm} \mathrm{g} \) So, 7,389.02 g of ethanol should be added to 620 kg of the paraffin to produce 1.07 kPa of ethanol vapor pressure at 35°C.