The vapour pressure of pure benzene at a cert | vedclass

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

Question

(C) Given: $p_{benzene}^o = 0.850 \ bar$,$p_{solution} = 0.845 \ bar$,$W_{benzene} = 39.0 \ g$,$M_{benzene} = 78 \ g/mol$,$w_{solid} = 0.5 \ g$.Using

Vedclass · Accepted Answer

$170$

Vedclass · Answer

(C) Given: $p_{benzene}^o = 0.850 \ bar$,$p_{solution} = 0.845 \ bar$,$W_{benzene} = 39.0 \ g$,$M_{benzene} = 78 \ g/mol$,$w_{solid} = 0.5 \ g$.Using Raoult's law for non-volatile solutes: $\frac{p^o - p}{p^o} = \frac{n_{solid}}{n_{solid} + n_{benzene}} \approx \frac{n_{solid}}{n_{benzene}}$ (for dilute solutions).$n_{benzene} = \frac{39.0 \ g}{78 \ g/mol} = 0.5 \ mol$.$\frac{0.850 - 0.845}{0.850} = \frac{0.5 / M_{solid}}{0.5}$.$\frac{0.005}{0.850} = \frac{0.5}{M_{solid} 	imes 0.5} = \frac{1}{M_{solid}}$.$M_{solid} = \frac{0.850}{0.005} = 170 \ g/mol$.

Similar Questions

Among the following substances,the lowest vapour pressure is exerted by

At a certain temperature,$5 \ g$ of a non-electrolyte solute is dissolved in $100 \ g$ of water. The vapour pressure of the resulting solution is $2985 \ N/m^2$,and the vapour pressure of pure water is $3000 \ N/m^2$. Calculate the molar mass of the solute.

The vapour pressure of a solvent decreases by $2.5 \ mm \ Hg$ by adding a solute. What is the mole fraction of solute? (Vapour pressure of pure solvent is $250 \ mm \ Hg$)

The vapour pressure of a solvent decreases by $10 \ mm \ Hg$ if the mole fraction of a non-volatile solute is $0.2$. Calculate the vapour pressure of the pure solvent. (in $mm \ Hg$)

Dry air is passed through a solution containing $10 \ g$ of solute in $90 \ g$ of water and then through pure water. The loss in weight of the solution is $2.5 \ g$ and the loss in weight of the solvent is $0.05 \ g$. The molecular weight of the solute is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module