Question

A certain mass of a substance, when dissolved in $100$ g C ${}_6$H${}_6$, lowers the freezing point by 1.28${}^o$C. The same mass of solute dissolved in $100$ g water lowers the freezing point by $1.40$${}^o$C. If the substance has a normal molecular weight in benzene and is completely ionized in water, thus, how many ions does it dissociate in water?

[$K$${}_f$ for H${}_2$O and C${}_6$H${}_6$ are $1.86$ and $5.12$ K kg mol${}^{-1}$]

• A. $3$ ions
• B. $4$ ions
• C. $5$ ions
• D. $6$ ions

Answer 1

The correct option is A $3$ ions

The expression for the depression in the freezing point and the molar mass of the solute is as shown below :

$\Delta T_f=\frac{K_f\times W_2}{M_2\times W_1}$

Substitute values for benzene in the above expression.

$1.28=\frac{5.12\times W_2}{M_2\times 100}$

$\frac{W_2}{25}=M_2$

Now substitute values for water in the above expression.

$1.40=\frac{1.86\times W_2}{M_2\times 100}$

$\frac{W_2}{75}=M_2$

Since same mass of substance is dissolved in benzene and water, $\frac{\frac{W_2}{25}}{\frac{W_2}{75}}=\frac{M_{2,benzene}}{M_{2,water}}$

Thus, $\frac{M_{2,benzene}}{M_{2,water}}=3$

Thus, one molecule of substance dissociates into 3 molecules in water.

Therefore, the option is A.