(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

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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?

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

The Van't Hoff factor of $BaCl_2$ at a concentration of $0.01 \ M$ is $1.98$. The percentage dissociation of $BaCl_2$ at this concentration will be: (in $\%$)

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The molecular mass of acetic acid $(CH_3COOH)$ dissolved in water is $60$,whereas when dissolved in benzene,it is $120$. This difference in behavior is because:

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When $0.0106 \text{ mole}$ of acetic acid is dissolved in $1 \text{ kg}$ of water, the observed freezing point depression is $0.0205 \text{ K}$. If the calculated freezing point depression is $0.0197 \text{ K}$, the Van't Hoff factor $(i)$ and the degree of dissociation $(\alpha)$ of acetic acid are respectively:

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$A$ compound $MX_2$ has observed and normal molecular masses $65.6$ and $164$ respectively. Calculate the percentage of ionization of $MX_2$ ............ $\%$

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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:

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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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The Van't Hoff factor of $BaCl_2$ at a concentration of $0.01 \ M$ is $1.98$. The percentage dissociation of $BaCl_2$ at this concentration will be: (in $\%$)

The molecular mass of acetic acid $(CH_3COOH)$ dissolved in water is $60$,whereas when dissolved in benzene,it is $120$. This difference in behavior is because:

$A$ compound $MX_2$ has observed and normal molecular masses $65.6$ and $164$ respectively. Calculate the percentage of ionization of $MX_2$ ............ $\%$

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:

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