Question

Assertion (A): Elevation in boiling point is a colligative property.
Reason (R): Elevation in boiling point is directly proportional to molarity

Everywhere answer is given both (A) and (R) are correct and (R) is correct explanation of (A). But how?? isn't elevation in boiling point directly proportional to molal concentration?? here it is molar concentration​

Answer 1

Answer:

A and R are correct explanation of A

Answer 2

Both (A) and (R) are correct and (R) is correct explanation of (A) as explained below.

Explanation:

The increase in the boiling point of a solvent on the addition of solute is referred to as boiling point elevation. It is given by the formula:

• $\Delta T_{b} = K_{b}. m$

where$\Delta T_{b}$ is the boiling point elevation, $K_{b}$ is the boiling point elevation constant and m is the molality of the solution.

The molality of solution can be defined as the total number of moles in 1 kilogram of solvent. It is expressed as:

• $m=\frac{W_{2} }{W_{1}\times Mm_{2} }$

where W₂ is the mass of solute, W₁ is the mass of solvent and Mm₂ is the molar mass of solute.

Substituting the expression of molality in boiling point elevation expression:

• $\Delta T_{b} = K_{b} \frac{W_{2} }{W_{1} \times Mm_{2} }$

• $Mm_{2} =\frac{K_{b} W_{2} }{W_{1} \Delta T_{b} }$

The number of moles of a solute can be given as:

• $n = \frac{W}{Mm}$

Therefore,

• $\frac{Mm_{2} }{W_{2} } =\frac{K_{b} }{W_{1} \Delta T_{b} }$

Taking reciprocal:

• $n_{2} = \frac{\Delta T_{b} . W_{1} }{K_{b} }$

• Thus, n₂ ∝ $\Delta T_{b}$

The molarity (M) of a solution can be defined as the total number of moles of solute (n₂) present in 1 liter (V₁) of solution. It is expressed as:

• $M=\frac{n_{2} }{V_{1} }$

Substituting the value of n₂ in the above equation.

• $M = \frac{\Delta T_{b}. W_{1} }{K_{b} V_{1} }$

Hence, $\Delta T_{b}$ ∝ M.