Question

A certain mass of a substance when dissolved in 100 g of ${{C}_{6}}{{H}_{6}}$, lowers the freezing point by ${{1.28}^{o}}C$. The same mass of solute when dissolved in 100 g of water, lowers the freezing point by ${{1.40}^{o}}C$. If the substance has normal molecular weight in benzene and is completely dissociated in water, into how many ions does it dissociate in water ? ${{K}_{f}}\text{ for }{{H}_{2}}O\text{ and }{{\text{C}}_{2}}{{H}_{6}}$ is 1.86 and 5.12 K Kg/mol respectively.

Answer 1

Hint: The depression in freezing point will be proportional to the molality of the substance. The number of ions can be found using the equation,
\[\Delta {{T}_{f}}={{K}_{f}}\times m\]
The molality found by applying the equation for benzene can be then used for water, to obtain the answer.

Complete step by step answer:
- The freezing point of a substance is the temperature at which the liquid freezes to form a solid.
- When a solute is added to a solution, it causes a depression in the freezing point.

- The depression in freezing point is given by the equation,
\[\Delta {{T}_{f}}={{K}_{f}}\times m\]
Where \[\Delta {{T}_{f}}\] is the depression in freezing point, ${{K}_{f}}$ is the freezing point depression constant, and m is the molality.

- We have to first find the molality by applying the values for benzene. From the above equation we can find the molality of benzene, given that
\[\Delta {{T}_{f}}={{1.28}^{0}}C,{{K}_{f}}=5.12\text{ }K\text{ }Kg/mol\] for benzene.

- Substituting the values, we get
\[m=\Delta {{T}_{f}}/{{K}_{f}}=1.28/5.12=0.25mol/kg\] for benzene.
- When ionization occurs, the equation of depression in freezing point changes to
\[\Delta {{T}_{f}}=i\times {{K}_{f}}\times m\]
Where i is the number of dissolved particles also called Van’t Hoff factor.

- We have,
\[\Delta {{T}_{f}}={{1.40}^{0}}C,{{K}_{f}} = 1.86K\text{ }Kg/mol,m=0.25mol/kg\]
- Now, we can substitute the values in the new equation to find the i value.
\[i=\Delta {{T}_{f}}/({{K}_{f}}\times m) = 140/(1.86\times 0.25)=3\]
- So, we can conclude that the substance dissociates into 3 ions in water.

Note: - It is important to note that the i factor comes into the equation only when a dissociation or association occurs. In all other cases, the i value is 1. The m in the equation is molality whose unit is mol/kg. It should not be confused with molarity or normality.