A solution is made by mixing 6.000 g of C₁₀H₈ in 90.00 g of C₆H₆. If the vapor pressure of pure C₆H₆ is 39.30 torr, what is the vapor pressure of the solution? (Molar masses: C₁₀H₈ = 128.0 g/mol; C₆H₆ = 78.00 g/mol)

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The vapor pressure of the solution can be calculated using Raoult's Law, which states that the vapor pressure of a solution is directly proportional to the mole fraction of the solvent. First, calculate the moles of each component:- Moles of $C_{10}H_{18}$ (solute): $\frac{6.000 \, \text{g}}{128.0 \, \text{g/mol}} = 0.046875 \, \text{mol}$- Moles of $C_{6}H_{6}$ (solvent): $\frac{90.00 \, \text{g}}{78.00 \, \text{g/mol}} = 1.1538 \, \text{mol}$Next, calculate the mole fraction of the solvent ($C_{6}H_{6}$):- Mole fraction of $C_{6}H_{6}$ = $\frac{1.1538}{1.1538 + 0.046875} = \frac{1.1538}{1.200675} \approx 0.961$Finally, calculate the vapor pressure of the solution:- Vapor pressure of the solution = Mole fraction of $C_{6}H_{6}$ $\times$ Vapor pressure of pure $C_{6}H_{6}$- Vapor pressure of the solution = $0.961 \times 39.30 \, \text{torr} \approx 37.76 \, \text{torr}$

A Solution Is Made by Mixing 6