If a solution prepared by adding $6.1 \ g$ of benzoic acid $(C_6H_5COOH)$ in $500 \ g$ of benzene freezes at $-0.290 \ ^oC$,find the percentage of association. (Given: $K_f$ of benzene $= 5.12 \ K \ kg/mol$,freezing point of pure benzene $= 5.50 \ ^oC$,molar mass of benzoic acid $= 122 \ g/mol$) (in $\%$)

  • A
    $90$
  • B
    $95$
  • C
    $80$
  • D
    $98$

Explore More

Similar Questions

In a solution of sulfur in carbon disulfide,$80 \%$ of the sulfur exists as $S_8$ and the remaining exists as $S_2$. The van't Hoff factor '$i$' will be:

The difference between the boiling point and the freezing point of a $0.2 \ m$ solution of acetic acid in benzene is $75.7 \ ^\circ C$. Calculate the value of the van't Hoff factor $i$. (For benzene,$K_b = 2.65 \ K \ m^{-1}$,$K_f = 5.12 \ K \ m^{-1}$,$T_b^o = 80 \ ^\circ C$,$T_f^o = 5.5 \ ^\circ C$)

$75.2 \ g$ of $C_6H_5OH$ (phenol) is dissolved in a solvent of $K_f = 14 \ K \ kg \ mol^{-1}$. If the depression in freezing point is $7 \ K$,then find the percentage of phenol that dimerises. (in $\%$)

Acetic acid dissolved in benzene shows a molecular mass of

Difficult
View Solution

Phenol associates in benzene to a certain extent to form a dimer. $A$ solution containing $20 \times 10^{-3} \ kg$ of phenol in $1 \ kg$ of benzene has its freezing point decreased by $0.69 \ K$. Calculate the percentage degree of association of phenol. ($K_f$ for benzene $= 5.12 \ K \ kg \ mol^{-1}$)

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo