Question

The $0.1$ molal aqueous solution of glucose boils at $100.16^\circ {\text{ C}}$. The boiling point of $0.5$ molal aqueous solution of sucrose will be:
A.$500.80^\circ {\text{ C}}$
B.$100.80^\circ {\text{ C}}$
C.$20.16^\circ {\text{ C}}$
D.$20.8^\circ {\text{ C}}$

Answer 1

The above question is based on colligative property known as elevation in boiling point. Using the formula we will first calculate the molal elevation constant and then using the formula to calculate the boiling point of the solution.
Formula used: $\Delta {{\text{T}}_{\text{b}}} = {{\text{K}}_{\text{b}}} \times {\text{m}}$
Here $\Delta {{\text{T}}_{\text{b}}}$ is elevation in boiling point, ${{\text{K}}_{\text{b}}}$ is molal elevation constant and m is the molality of the solution.

Complete step by step answer:
The molality of the aqueous solution of glucose is given to us. We know the normal boiling point of water is hundred degrees Celsius but after the addition of glucose the boiling point becomes $101.16^\circ {\text{ C}}$ as given to us. So the elevation in boiling point will be $\left( {100.16 - 100} \right)^\circ {\text{C }} = 0.16\,^\circ {\text{C}}$
Using the formula we will first calculate the molar elevation constant for the given data which will be equal to: $\dfrac{{\Delta {{\text{T}}_{\text{b}}}}}{{\text{m}}} = {{\text{K}}_{\text{b}}}$
${{\text{K}}_{\text{b}}} = \dfrac{{0.16\,^\circ {\text{C}}}}{{0.1{\text{ m}}}} = 1.6{\text{ }}^\circ {\text{C}}/{\text{m}}$
Since model elevation constant will be a constant quantity as it depends on the temperature and not on the concentration of solution. So for 0.5 molal aqueous solution, the molal elevation constant will be the same. So using the data we will calculate the elevation in boiling point.
$\Delta {{\text{T}}_{\text{b}}} = 1.6 \times 0.5 = 0.80$
This is the elevation in boiling point but we need to calculate the boiling point of solution.
${\text{ boiling point of solution }} - {\text{ boiling point of solvent }} = {\text{ }}\Delta {{\text{T}}_{\text{b}}}$
So the boiling point of the solution will be: $(0.80 + 100)^\circ {\text{C }} = 100.80^\circ {\text{C}}$.

Hence the correct option is B.

Note:
Colligative properties are those properties which depend upon the number of moles or amount of solute present it in the solution. As we can see that when we increase the amount of solute the elevation in boiling point also increases. This means that elevation in boiling point is a colligative property.