Question

On mixing $ 100 $ ml of $ 2 $ M $ NaCl $ and $ 100 $ ml of $ 2 $ M $ CaC{l_2} $ . What is elevation in boiling point in $ ^0C $ $ \left[ {{K_b}\left( {{H_2}O} \right) = 0.2Kgmo{l^{ - 1}}} \right] $ (assume molarity is equal to molality).

Answer 1

Hint: The elevation in boiling point of a solvent is due to the addition of a solute. The boiling increases due to the addition of a solute. Given that the molarity and molality are equal. Thus, molality can be calculated from molarity. It can be calculated from the molality, van’t Hoff factor, and molar elevation constant.
 $ \Delta {T_b} = {K_b} \times m \times i $
Where,
 $ \Delta {T_b} $ is elevation in boiling point
 $ {K_b} $ is molar elevation constant
M is molality or molarity (molarity is equal to molality)
I, is the Van't Hoff factor.

Complete answer:
Given that $ 100 $ ml of $ 2 $ M $ NaCl $ and $ 100 $ ml of $ 2 $ M $ CaC{l_2} $ is mixed.
The molarity can be calculated from number of moles of solute and volume of solution in litres
Number of moles of $ NaCl $ is equal to $ 2 \times 0.1 = 0.2 $
Number of moles of $ CaC{l_2} $ is equal to $ 2 \times 0.1 = 0.2 $
Thus, the mixture consists of total number of moles will be $ 0.2 + 0.2 = 0.4 $
The total volume will be equal to $ 200 $ ml
Thus, the molarity will be $ \dfrac{{0.4}}{{0.2}} = 2M $
As the molarity and molality are same, the molality will be equal to $ 2m $
The elevation in boiling point will be calculated by substituting all the above values
 $ \Delta {T_b} = 0.2 \times 2 \times 3 = {1.2^0}C $
Thus, the elevation in boiling point is $ {1.2^0}C $ .

Note:
Since the two solutions added were sodium chloride and calcium chloride. The common ion in the two solutions was chloride ion. The sodium ions, calcium ions and chloride ions were formed. Thus, the total ions formed were $ 3 $ . Thus, the van’t Hoff factor value will be $ 3 $ .