Question

biphenyl, $c_{12}h_{10}$, is a nonvolatile, nonionizing solute that is soluble in benzene, $c_{6}h_{6}$. at 25 $^{circ}c$, the vapor - pressure of pure benzene is 100.84 torr. what is the vapor pressure of a solution made from dissolving 15.3 g of biphenyl in 27.3 g of benzene?

Explanation:

Step1: Calculate moles of biphenyl

The molar - mass of biphenyl ($C_{12}H_{10}$) is $M_{C_{12}H_{10}}=(12\times12 + 1\times10)\ g/mol=154\ g/mol$.
The number of moles of biphenyl, $n_{C_{12}H_{10}}=\frac{m_{C_{12}H_{10}}}{M_{C_{12}H_{10}}}=\frac{15.3\ g}{154\ g/mol}=0.09948\ mol$.

Step2: Calculate moles of benzene

The molar - mass of benzene ($C_{6}H_{6}$) is $M_{C_{6}H_{6}}=(12\times6 + 1\times6)\ g/mol = 78\ g/mol$.
The number of moles of benzene, $n_{C_{6}H_{6}}=\frac{m_{C_{6}H_{6}}}{M_{C_{6}H_{6}}}=\frac{27.3\ g}{78\ g/mol}=0.35\ mol$.

Step3: Calculate mole fraction of benzene

The mole fraction of benzene, $x_{C_{6}H_{6}}=\frac{n_{C_{6}H_{6}}}{n_{C_{6}H_{6}}+n_{C_{12}H_{10}}}=\frac{0.35\ mol}{0.35\ mol + 0.09948\ mol}=\frac{0.35\ mol}{0.44948\ mol}=0.779$.

Step4: Calculate vapor pressure of the solution

According to Raoult's law, $P = x_{solvent}P_{solvent}^0$.
Given $P_{C_{6}H_{6}}^0 = 100.84\ Torr$ and $x_{C_{6}H_{6}} = 0.779$.
$P=(0.779)\times100.84\ Torr = 78.65\ Torr$.

Answer:

$78.65\ Torr$