Chemistry - Liquid Solution Question with Solution | TestHub

The colligative property of freezing point depression is given by the formula:

$$\Delta T_f=i{K}_{f}m$$

where $\Delta T_f$ is the depression in freezing point, $i$ is the van't Hoff factor, $K_f$ is the cryoscopic constant, and $m$ is the molality of the solution.

 

In benzene (C₆H₆), the substance has normal molecular mass, meaning it does not associate or dissociate. Thus, the van't Hoff factor $i=1$.

Given $\Delta T_f=1.28$

 

and $K_f=5.12Kkgmo{l}^{-1}$.

$$1.28=(1)\times 5.12\times m_{benzene}$$

$$m_{benzene}=\frac{1.28}{5.12}=0.25molk{g}^{-1}$$

Since the mass of solvent is $100g=0.1kg$, the moles of solute are $0.25molk{g}^{-1}\times 0.1kg=0.025mol$.

 

In water, the same mass of solute is dissolved in $100g$ of water. The molality of the solution in water is also $0.25molk{g}^{-1}$ (since the mass of solute and solvent are the same as in benzene).

Given $\Delta T_f=1.40^{}$

 

and $K_f=1.86Kkgmo{l}^{-1}$ for water.

$$1.40=i_{water}\times 1.86\times 0.25$$

$$i_{water}=\frac{1.40}{1.86\times 0.25}=\frac{1.40}{0.465}\approx 3.01$$

Since the substance is completely dissociated in water, the van't Hoff factor $i$ represents the number of ions into which one molecule of the substance dissociates.

Therefore, the substance dissociates into approximately 3 ions in water.

 

The final answer is $\boxed{3}$.