(D) $\Delta T_f = T_o - T_f = 5.3 - 4.47 = 0.83 \ K$ Molar mass of $C_6H_5CH_2COOH = 136 \ g/mol$ Molality $(m) = \frac{0.223 \ g}{136 \ g/mol} \times

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Phenylacetic acid forms a dimer in benzene.

(D) $\Delta T_f = T_o - T_f = 5.3 - 4.47 = 0.83 \ K$ Molar mass of $C_6H_5CH_2COOH = 136 \ g/mol$ Molality $(m) = \frac{0.223 \ g}{136 \ g/mol} \times \frac{1000}{4.49 \ g} = 0.365 \ mol/kg$ Using $\Delta T_f = i \times K_f \times m$: $i = \frac{\Delta T_f}{K_f \times m} = \frac{0.83}{5.12 \times 0.365} \approx 0.444$ Since $i \approx 0.5$,it indicates that the solute molecules associate to form a dimer in benzene.

Class 12 Chemistry Solutions Abnormal molecular mass

Medium

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
Phenylacetic acid $(C_6H_5CH_2COOH)$ remains as a monomer in benzene.
B
Phenylacetic acid undergoes partial ionization in benzene.
C
Phenylacetic acid undergoes complete ionization in benzene.
D
Phenylacetic acid forms a dimer in benzene.

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Abnormal molecular mass

If a $0.02 \, M \, Pb(NO_3)_2$ solution is isotonic with a $0.05 \, M$ aqueous glucose solution,the degree of dissociation of $Pb(NO_3)_2$ will be ..........

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Three particles of a solute $A$ associate in benzene to form a trimer $A_3$. Calculate the freezing point of a $0.25 \ m$ solution if the degree of association of solute $A$ is $0.80$. The freezing point of benzene is $5.5 \ ^oC$ and $K_f$ is $5.12 \ K \ kg \ mol^{-1}$. (Answer in $K$)

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The degree of dissociation $(\alpha)$ of a weak electrolyte,$A_xB_y$ is related to van't Hoff factor $(i)$ by the expression

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The freezing point of a $0.262 \ mol \ kg^{-1}$ solution of acetic acid in benzene is $277.4 \ K$. If the $K_f$ value for benzene is $5.0 \ K \ kg \ mol^{-1}$ and the freezing point of pure benzene is $278.4 \ K$,then the van't Hoff factor is ...........

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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,$K_b = 2.65 \ K \ m^{-1}$,$K_f = 5.12 \ K \ m^{-1}$,$T_b^o = 80 \ ^\circ C$,$T_f^o = 5.5 \ ^\circ C$)

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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?

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If a $0.02 \, M \, Pb(NO_3)_2$ solution is isotonic with a $0.05 \, M$ aqueous glucose solution,the degree of dissociation of $Pb(NO_3)_2$ will be ..........

Three particles of a solute $A$ associate in benzene to form a trimer $A_3$. Calculate the freezing point of a $0.25 \ m$ solution if the degree of association of solute $A$ is $0.80$. The freezing point of benzene is $5.5 \ ^oC$ and $K_f$ is $5.12 \ K \ kg \ mol^{-1}$. (Answer in $K$)

The degree of dissociation $(\alpha)$ of a weak electrolyte,$A_xB_y$ is related to van't Hoff factor $(i)$ by the expression

The freezing point of a $0.262 \ mol \ kg^{-1}$ solution of acetic acid in benzene is $277.4 \ K$. If the $K_f$ value for benzene is $5.0 \ K \ kg \ mol^{-1}$ and the freezing point of pure benzene is $278.4 \ K$,then the van't Hoff factor is ...........

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