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The vapour pressures of benzene and tolu...

The vapour pressures of benzene and toluene at 293 K are 75 mm and 22 mm Hg respectively. 23.4g of benzene and 64.4 g of toluene are mixed. If the two form and ideal solution, calculate the mole fraction of benzene in the vapour phase assuming that the vapours are in equilibrium with the liquid mixture at this temperature.

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To solve the problem step by step, we will follow these calculations: ### Step 1: Calculate the moles of benzene and toluene. 1. **Moles of Benzene (C6H6)**: - Given mass of benzene = 23.4 g - Molar mass of benzene = 78 g/mol (calculated as \(6 \times 12 + 6 \times 1\)) - Moles of benzene (Na) = \(\frac{\text{mass}}{\text{molar mass}} = \frac{23.4 \, \text{g}}{78 \, \text{g/mol}} = 0.3 \, \text{mol}\) ...
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