Question

A solution containing 7.5 g of urea (molecular mass = 60) in 1 kg of water freezes at the same temperature as another solution containing 15 g of solute 'A’ in the same amount of water. Calculate molar mass of 'A.
A.120
B.20
C.110
D.60

Answer 1

Depression in freezing point: The property of decrease in freezing point when some non-volatile solute is dissolved is called depression of freezing point. The depression in freezing point is represented as $\mathop {\Delta T}\nolimits_f$.
Freezing Point: Temperature at which the liquid and the solid forms of the solvent are in equilibrium and hence have the same vapour pressure is called freezing point.

Complete step by step answer:
$\mathop {\Delta T}\nolimits_f \propto m$(where m is the molarity)
No of moles of urea $\mathop n\nolimits_1 = {\text{ }}\dfrac{{7.5}}{{60}}{\text{ }} = {\text{ }}0.125$
No. Of moles of solute $A{\text{ }} = {\text{ }}\dfrac{{15}}{{\mathop M\nolimits_2 }}$( where $\mathop M\nolimits_2$ is molar mass of solute)
As the freezing point is same they will have same number of moles due to same molality:
$\dfrac{{7.5}}{{60}} = \dfrac{{15}}{{\mathop M\nolimits_2 }}$
Now $\mathop M\nolimits_2 {\text{ }} = {\text{ }}120g$
Our required answer is A that is 120.

Note:
Properties of solutions depending only upon the number of solute particles per unit volume are known as colligative properties. They are
1.Relative lowering of vapour pressure
2.Elevation in boiling points
3.Depression in freezing point
4.Osmotic pressure
These properties are valid and applicable only for dilute solutions in which the solute is a non-volatile solid.