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Question: A certain mass of a substance when dissolved in 100 g of \({{C}_{6}}{{H}_{6}}\), lowers the freezing...

A certain mass of a substance when dissolved in 100 g of C6H6{{C}_{6}}{{H}_{6}}, lowers the freezing point by 1.28oC{{1.28}^{o}}C. The same mass of solute when dissolved in 100 g of water, lowers the freezing point by 1.40oC{{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 ? Kf for H2O and C2H6{{K}_{f}}\text{ for }{{H}_{2}}O\text{ and }{{\text{C}}_{2}}{{H}_{6}} is 1.86 and 5.12 K Kg/mol respectively.

Explanation

Solution

The depression in freezing point will be proportional to the molality of the substance. The number of ions can be found using the equation,
ΔTf=Kf×m\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,
ΔTf=Kf×m\Delta {{T}_{f}}={{K}_{f}}\times m
Where ΔTf\Delta {{T}_{f}} is the depression in freezing point, Kf{{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
ΔTf=1.280C,Kf=5.12 K Kg/mol\Delta {{T}_{f}}={{1.28}^{0}}C,{{K}_{f}}=5.12\text{ }K\text{ }Kg/mol for benzene.

- Substituting the values, we get
m=ΔTf/Kf=1.28/5.12=0.25mol/kgm=\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
ΔTf=i×Kf×m\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,
ΔTf=1.400C,Kf=1.86K Kg/mol,m=0.25mol/kg\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=ΔTf/(Kf×m)=140/(1.86×0.25)=3i=\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.