Question

The molal boiling point constant of water is ${0.573^o}Ckgmo{l^{ - 1}}$ . When $0.1$ mole of glucose is dissolved in $1000g$ of water, the solution boils under atmospheric pressure at:
(A) ${100.513^o}C$
(B) ${100.0573^o}C$
(C) ${100.256^o}C$
(D) ${101.025^o}C$

Answer 1

Hint: Colligative properties are the properties of a liquid which depends upon the amount (number of moles) of the solute added to it. One such property is elevation in boiling point which states that when a non-volatile solute is added to the solution, boiling point of solution increases. So, in the solution we will see the formula to calculate the elevation in boiling point of the solution containing non-volatile solute.

Complete answer:
 We know that colligative properties are those properties which depend upon the amount of the solute in the given solution.
One such property is elevation in boiling point which states that when a non-volatile solute is added to the solution, boiling point of solution increases.
Formula for the elevation in boiling point is given as:
$\Delta {T_b} = {K_b}m$
Here, ${K_b} = {0.573^o}Ckgmo{l^{ - 1}}$
$n = 0.1mole$
M is mass of solvent which is equals to $1000g$ .
$M = 1kg$
Formula for the elevation in boiling point can be rewritten as:
$\Delta {T_b} = \dfrac{{{K_b} \times n}}{{M(kg)}}$
So we now insert given values in above equation, we get
$\Delta {T_b} = \dfrac{{0.573 \times 0.1}}{1}$
$\Delta {T_b} = 0.0573$
Therefore, the boiling point of solution can be found as:
$100 + 0.0573 = {100.0573^o}C$
Hence, the correct option is (B) ${100.0573^o}C$ .

Note:
We should note that while calculating the molality of the solution, the mass of solvent is taken into account and the mass should be in kilogram units. We have added the elevation in boiling point to the boiling point of pure water because we need to find the boiling point of the whole solution when solute is added to it.