Question

If 0.15g of a solute dissolved in 15 g of solvent is boiled at a temperature higher by 0.216$^{o}C$ than that of the pure solvent, then the molecular weight of the substance is (molal elevation constant for the solvent is 2.16):
(A) 1.01
(B) 10
(C) 10.1
(D) 100

Answer 1

Hint: When a solute is dissolved in a solvent, the boiling point decreases. Thus, the boiling point of a solution is always greater than the boiling point of a pure solvent. This rise in boiling point is referred to as elevation of boiling point, and it is denoted by $\Delta {{T}_{b}}$. The elevation of the boiling point is directly proportional to the molal concentration of the solute in solution ($\Delta {{T}_{b}}\propto m$).

Formula Used: The molecular weight of a substance ($m$) can be calculated as:
$m=\frac{{{K}_{b}}\times w\times 1000}{\Delta {{T}_{b}}\times W}$; where ${{K}_{b}}$= Molal elevation constant, $w$= Weight of solute, $W$ = Weight of solvent, $\Delta {{T}_{b}}$ = Elevation of boiling point

Complete Step by Step Answer:
The molecular weight of a substance can be calculated as:
$m=\frac{{{K}_{b}}\times w\times 1000}{\Delta {{T}_{b}}\times W}$
$m=\frac{2.16\times 0.15\times 1000}{0.216\times 15}$
$m=100$

Knowing the molecular weight ($m$), solute weight ($w$), and solvent weight ($W$), the elevation of boiling point ($\Delta {{T}_{b}}$) can also be calculated for a solution whose molal elevation constant value (${{K}_{b}}$) is known.
Correct Option: (D) 100.

Note: The units of molal elevation constant (${{K}_{b}}$) are $K$$kg$$mo{{l}^{-1}}$. The colligative properties like elevation of the boiling point of the solvent, depression of the freezing point of the solvent, relative lowering of the vapour pressure of the solvent and osmotic pressure of a solution depend upon the number of solute particles with respect to the total number of particles present in the solution.