Chemistry - Liquid Solution Question with Solution | TestHub

Total vapour pressure of mixture $=$ (Mole fraction of pentane $\times \mathrm{VP}$ of pentane $)$ + (Mole fraction of hexane $\times \mathrm{VP}$ of hexane) $=\mathrm{VP}$ of pentane in mixture +VP of hexane in mixture $=\left(\frac{1}{5}\times 440+\frac{4}{5}\times 120\right)=184\mathrm{mm}$ $\because$ VP of Pentane in mixture. $=\mathrm{VP}$ of mixture $\times$ mole fraction of pentane in vapour phase $88=184\times$ mole fraction of pentane in vapour phase $\therefore$ Mole fraction of pentane in vapour phase $=\frac{88}{184}=0.478$