When 12.2 g of benzoic acid is dissolved in | vedclass

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

Question

(C) The freezing point depression is given by $\Delta T_{f} = i 	imes K_{f} 	imes m$. Given $\Delta T_{f} = 0 - (-0.93) = 0.93 \ K$. Molality $m = \

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) The freezing point depression is given by $\Delta T_{f} = i 	imes K_{f} 	imes m$. Given $\Delta T_{f} = 0 - (-0.93) = 0.93 \ K$. Molality $m = \frac{	ext{mass of solute}}{	ext{molar mass of solute}} 	imes \frac{1000}{	ext{mass of solvent in g}} = \frac{12.2}{122} 	imes \frac{1000}{100} = 1 \ m$. Substituting values: $0.93 = i 	imes 1.86 	imes 1$,which gives $i = 0.5$. For association,the van't Hoff factor is $i = 1 + (\frac{1}{n} - 1)\alpha$. Since $\alpha = 1$ ($100 \%$ association),$0.5 = 1 + (\frac{1}{n} - 1) 	imes 1$. $0.5 = \frac{1}{n}$,therefore $n = 2$.

When $12.2 \ g$ of benzoic acid is dissolved in $100 \ g$ of water,the freezing point of the solution was found to be $-0.93^{\circ} C$ $(K_{f}(H_{2}O) = 1.86 \ K \ kg \ mol^{-1})$. The number $(n)$ of benzoic acid molecules associated (assuming $100 \ \%$ association) is ........ .

Similar Questions

The experimental molecular mass of acetic acid dissolved in benzene will be ........ (Assume complete association).

$A$ solute $A$ undergoes association in a medium as $nA \rightleftharpoons A_n$. If the degree of association is $\alpha$,then the van't Hoff factor $i$ is given by:

Determination of the molar mass of acetic acid in benzene using freezing point depression is affected by

What is abnormal molar mass?

$A$ $0.004 \ M$ 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 $..... \ \%$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module