Question

Which of the following will produce the ebullioscopic constant which is the elevation of the boiling point?
A.1 molal solution
B.1 molar solution
C.1 Normal solution
D.$100\%$ solution

Answer 1

Certain properties of solutions depend upon the number of solute particles (present in the form of molecules or ions), present in a definite amount of solvent, but not on the chemical nature of the solute is known as colligative properties.

Complete step by step answer:
Step 1
The four important colligative properties of a solution are as follows:
Relative lowering of vapor pressure
The elevation of boiling point
The depression of freezing point
Osmotic pressure
Step 2
The temperature at which the equilibrium vapor pressure of a liquid becomes equal to the atmospheric pressure is known as the boiling point of that liquid.
Step 3
When a non-volatile solute is dissolved in a solvent, its boiling point is raised. The increase in the boiling point is referred to as the elevation of the boiling point. The elevation of the boiling point is directly proportional to the relative lowering in vapor pressure.
$\Delta {T_b}\alpha \Delta p$
$\Delta {T_b} =$ The elevation in the boiling point on the addition of non-volatile solute and $\Delta p$ is the relative lowering of the vapour pressure.
Step 4
The mathematical expression for the Raoult’s law, for a dilute solution,
$\dfrac{p^0 - p}{p^0} = \dfrac{w}{M'} \times \dfrac{M}{W}$
where $\dfrac{p^0 - p}{p^0}$ is the relative lowering of vapor pressure, w is the mass of the solute, M’ is the molecular mass of the solute, W is the mass of the solvent, the molecular mass of solvent is denoted by M.
$\dfrac{\Delta p}{p^0} = \dfrac{w}{M'} \times \dfrac{M}{W}$
$\Delta p = {p^0}M \times \dfrac{w}{WM'}$
As, ${p^0}M$ is constant for a particular solvent, because vapour pressure of the pure solvent at its boiling point and the molecular mass of the solvent both are constant terms.
So, $\Delta p\alpha \dfrac{w}{WM'}$
$\Delta {T_b}\alpha \dfrac{w}{WM'}$
If the mass of the solvent is 1kg, $\dfrac{w}{WM'}$ represents the molality of the solution.
Therefore,
$\Delta {T_b}\alpha m$
$\Delta {T_b} = {K_b}m$
Here ${K_b}$ is the proportionality constant which is also known as the ebullioscopic constant.
This represents a relationship between the elevation of the boiling point and the molality of the solution. In this mathematical expression, Kb represents the molal elevation constant which is also known as the ebullioscopic constant.
So, you can find that the elevation of boiling point of a liquid in addition to a non-volatile solid is dependent on the molality of the solution.

Hence, option a) 1 molal solution will produce the ebullioscopic constant which is the molal elevation constant is the correct answer for the given question.

Note:
The depression in the freezing point also depends upon the molality of the solution and molal depression constant is also known as the cryoscopic constant.