75.2 g of C_6H_5OH (phenol) is dissolved in | vedclass

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

Question

(B) The molar mass of phenol $(C_6H_5OH)$ is $M = 94 \ g \ mol^{-1}$.Given mass of phenol $w = 75.2 \ g$,so moles $n = \frac{75.2}{94} = 0.8 \ mol$.As

Vedclass · Accepted Answer

$75$

Vedclass · Answer

(B) The molar mass of phenol $(C_6H_5OH)$ is $M = 94 \ g \ mol^{-1}$.Given mass of phenol $w = 75.2 \ g$,so moles $n = \frac{75.2}{94} = 0.8 \ mol$.Assuming $1 \ kg$ of solvent,molality $m = 0.8 \ mol \ kg^{-1}$.The formula for depression in freezing point is $\Delta T_f = i 	imes K_f 	imes m$.Given $\Delta T_f = 7 \ K$ and $K_f = 14 \ K \ kg \ mol^{-1}$,we have $7 = i 	imes 14 	imes 0.8$.$i = \frac{7}{14 	imes 0.8} = \frac{1}{1.6} = 0.625$.For dimerization,$i = 1 - \alpha + \frac{\alpha}{2} = 1 - \frac{\alpha}{2}$.$0.625 = 1 - \frac{\alpha}{2} \implies \frac{\alpha}{2} = 0.375 \implies \alpha = 0.75$.Thus,the percentage of phenol that dimerises is $75\%$.

$75.2 \ g$ 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 $\%$)

Similar Questions

If $0.1 \ M$ solution of $NaCl$ is isotonic with $1.1 \ w \%$ urea solution, the degree of ionisation of $NaCl$ is $($Molar masses of urea and $NaCl$ are $60 \ g \ mol^{-1}$ and $58.5 \ g \ mol^{-1}$, respectively.$)$

$0.004 \ M \ Na_2SO_4$ is isotonic with $0.01 \ M$ glucose. The degree of dissociation of $Na_2SO_4$ is ........... $\%$.

$1.2 \ g$ of $NaCl$ solution is isotonic with $7.2 \ g$ of glucose solution. Calculate the van’t Hoff factor $(i)$ of the $NaCl$ solution.

The experimental depression in freezing point of a dilute solution is $0.025 \ K$. If the van't Hoff factor $(i)$ is $2.0$, the calculated depression in freezing point (in $K$) is

If a $0.01 \ m$ aqueous solution of an electrolyte freezes at $-0.056 \ ^{\circ}C$,calculate the van't Hoff factor $(i)$ for the electrolyte. (Cryoscopic constant of water $K_{f} = 1.86 \ K \ kg \ mol^{-1}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module