The vapour pressure of pure benzene and tolue | vedclass

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

Question

(A) $1$. Molar mass of benzene $(C_6H_6)$ = $78 \ g/mol$. Molar mass of toluene $(C_7H_8)$ = $92 \ g/mol$. $2$. Moles of benzene $(n_B)$ = $7.8 \ g /

Vedclass · Accepted Answer

$0.854$

Vedclass · Answer

(A) $1$. Molar mass of benzene $(C_6H_6)$ = $78 \ g/mol$. Molar mass of toluene $(C_7H_8)$ = $92 \ g/mol$. $2$. Moles of benzene $(n_B)$ = $7.8 \ g / 78 \ g/mol = 0.1 \ mol$. $3$. Moles of toluene $(n_T)$ = $180 \ g / 92 \ g/mol \approx 1.9565 \ mol$. $4$. Mole fraction of benzene $(x_B)$ = $0.1 / (0.1 + 1.9565) = 0.1 / 2.0565 \approx 0.0486$. $5$. Mole fraction of toluene $(x_T)$ = $1 - 0.0486 = 0.9514$. $6$. Total vapour pressure $(P_{total})$ = $P_B^0 	imes x_B + P_T^0 	imes x_T$. $7$. $P_{total} = (0.9 	imes 0.0486) + (0.85 	imes 0.9514) = 0.04374 + 0.80869 = 0.85243 \ Bar$.

The vapour pressure of pure benzene and toluene are $0.9 \ Bar$ and $0.85 \ Bar$ respectively. Calculate the vapour pressure of the solution prepared by dissolving $7.8 \ g$ of benzene in $180 \ g$ of toluene. (in $Bar$)

Similar Questions

The vapour pressure of a dilute aqueous solution of glucose is $750 \ mm$ of mercury at $373 \ K$. The mole fraction of solvent is

The vapour pressure of water at $23^{\circ} C$ is $19.8 \ mm$. If $0.1 \ mole$ of glucose is dissolved in $178.2 \ g$ of water, what is the vapour pressure (in $mm$) of the resultant solution?

What is the vapour pressure of a solution containing $0.1 \ mol$ of solute dissolved in $1.8 \times 10^{-2} \ kg$ of $H_2O$ (in $mm \ Hg$)? $(P_1^0 = 24 \ mm \ Hg)$

The graph between vapour pressure and mole fraction of the solvent for an ideal solution is:

At the same temperature,which of the following will have the highest vapor pressure?

Vedclass Test Series

Exam Paper Generator

Online Exam Module