Phenol associates in benzene to a certain ext | vedclass

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

Question

(C) The molar mass of phenol $(C_6H_5OH)$ is $94 \ g \ mol^{-1}$.First,calculate the molality $(m)$ of the solution:$m = \frac{	ext{mass of solute (g

Vedclass · Accepted Answer

$73.3$

Vedclass · Answer

(C) The molar mass of phenol $(C_6H_5OH)$ is $94 \ g \ mol^{-1}$.First,calculate the molality $(m)$ of the solution:$m = \frac{	ext{mass of solute (g)}}{	ext{molar mass} 	imes 	ext{mass of solvent (kg)}} = \frac{20 \ g}{94 \ g \ mol^{-1} 	imes 1 \ kg} = 0.2128 \ mol \ kg^{-1}$.Using the freezing point depression formula: $\Delta T_f = i 	imes K_f 	imes m$.$0.69 = i 	imes 5.12 	imes 0.2128$.$i = \frac{0.69}{1.0895} \approx 0.633$.For the association of phenol into a dimer $(2C_6H_5OH ightleftharpoons (C_6H_5OH)_2)$,the van't Hoff factor is $i = 1 - \alpha + \frac{\alpha}{2} = 1 - \frac{\alpha}{2}$.$0.633 = 1 - \frac{\alpha}{2}$.$\frac{\alpha}{2} = 1 - 0.633 = 0.367$.$\alpha = 0.734$,which is $73.4 \%$. The closest option is $73.3 \%$.

Similar Questions

Pure benzene freezes at $5.3^oC$. $A$ solution of $0.223 \ g$ of $C_6H_5CH_2COOH$ in $4.49 \ g$ of benzene freezes at $4.47^oC$. Given $K_f = 5.12 \ K \ kg/mol$,what can be concluded from this?

$A$ $0.004 \ M$ solution of $Na_2SO_4$ is isotonic with a $0.010 \ M$ solution of glucose at the same temperature. The apparent degree of dissociation of $Na_2SO_4$ is $..... \ \%$.

The observed osmotic pressure of a solution of benzoic acid in benzene is less than its expected value because:

One molal solution of a carboxylic acid in benzene shows the elevation of boiling point of $1.518 \ K$. The degree of association for dimerization of the acid in benzene is $........ \%$. ($K_b$ for benzene $= 2.53 \ K \ kg \ mol^{-1}$)

$HA_{(aq)} \rightleftharpoons H^{+}_{(aq)} + A^{-}_{(aq)}$
The freezing point depression of a $0.1 \ m$ aqueous solution of a monobasic weak acid $HA$ is $0.20^{\circ} C$. The dissociation constant for the acid is. Given: $K_{f}(H_2O) = 1.8 \ K \ kg \ mol^{-1}$,molality $\equiv$ molarity.

Vedclass Test Series

Exam Paper Generator

Online Exam Module