The vapour pressures of chloroform (CHCl_3) a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) $1$. Calculate moles of each component: $n_{CHCl_3} = \frac{59.75 \ g}{119.5 \ g \cdot mol^{-1}} = 0.5 \ mol$ $n_{CH_2Cl_2} = \frac{21.25 \ g}{85

Vedclass · Accepted Answer

$0.491, 0.509$

Vedclass · Answer

(B) $1$. Calculate moles of each component: $n_{CHCl_3} = \frac{59.75 \ g}{119.5 \ g \cdot mol^{-1}} = 0.5 \ mol$ $n_{CH_2Cl_2} = \frac{21.25 \ g}{85 \ g \cdot mol^{-1}} = 0.25 \ mol$ $2$. Calculate mole fractions in liquid phase: $x_{CHCl_3} = \frac{0.5}{0.5 + 0.25} = \frac{0.5}{0.75} = \frac{2}{3}$ $x_{CH_2Cl_2} = \frac{0.25}{0.5 + 0.25} = \frac{0.25}{0.75} = \frac{1}{3}$ $3$. Calculate partial pressures: $P_{CHCl_3} = x_{CHCl_3} 	imes P^0_{CHCl_3} = \frac{2}{3} 	imes 200 \ mmHg = 133.33 \ mmHg$ $P_{CH_2Cl_2} = x_{CH_2Cl_2} 	imes P^0_{CH_2Cl_2} = \frac{1}{3} 	imes 415 \ mmHg = 138.33 \ mmHg$ $4$. Calculate total pressure: $P_{total} = 133.33 + 138.33 = 271.66 \ mmHg$ $5$. Calculate mole fractions in vapour phase $(y_i = \frac{P_i}{P_{total}})$: $y_{CHCl_3} = \frac{133.33}{271.66} \approx 0.491$ $y_{CH_2Cl_2} = \frac{138.33}{271.66} \approx 0.509$

Similar Questions

For an ideal solution,the correct option is

Which of the following pairs of liquids shows a positive deviation from Raoult's law?

Which of the following is not equal to zero for an ideal solution?

Vedclass Test Series

Exam Paper Generator

Online Exam Module