JEE MAIN - Chemistry (2023 - 8th April Evening Shift - No. 17)
If the boiling points of two solvents X and Y (having same molecular weights) are in the ratio $$2: 1$$ and their enthalpy of vaporizations are in the ratio $$1: 2$$, then the boiling point elevation constant of $$\mathrm{X}$$ is $$\underline{\mathrm{m}}$$ times the boiling point elevation constant of Y. The value of m is ____________ (nearest integer)
Answer
8
Explanation
The boiling point elevation constant, also known as the ebullioscopic constant ($K_{b}$), can be determined using the formula:
$$K_{b} = \frac{R \cdot T_{b}^{2}}{1000 \cdot \Delta H_{vap}}$$
where:
- $R$ is the universal gas constant (8.31 J mol⁻¹ K⁻¹)
- $T_{b}$ is the boiling point of the solvent (in K)
- $\Delta H_{vap}$ is the enthalpy of vaporization (in J/mol)
According to the problem, the ratio of boiling points for X and Y is 2:1, and the ratio of their enthalpy of vaporization is 1:2.
Let's denote the boiling point of Y as $T_{b}(Y)$ and the boiling point of X as $T_{b}(X)$, and likewise for the enthalpy of vaporization $\Delta H_{vap}(Y)$ and $\Delta H_{vap}(X)$.
We then have:
$T_{b}(X) = 2T_{b}(Y)$ and $\Delta H_{vap}(X) = 0.5\Delta H_{vap}(Y)$
We can substitute these values into the equation for $K_{b}$:
$K_{b}(X) = R*(2T_{b}(Y))^2 / (1000 * 0.5\Delta H_{vap}(Y)) = 4*R*T_{b}(Y)^2 / (500 * \Delta H_{vap}(Y))$
and
$K_{b}(Y) = R*T_{b}(Y)^2 / (1000 * \Delta H_{vap}(Y))$
Comparing these two equations:
$K_{b}(X) / K_{b}(Y) = [4*R*T_{b}(Y)^2 / (500 * \Delta H_{vap}(Y))] / [R*T_{b}(Y)^2 / (1000 * \Delta H_{vap}(Y))] $
This simplifies to:
$K_{b}(X) / K_{b}(Y) = 4 / 0.5 = 8$
So, the boiling point elevation constant of X is 8 times the boiling point elevation constant of Y. Therefore, m = 8.
$$K_{b} = \frac{R \cdot T_{b}^{2}}{1000 \cdot \Delta H_{vap}}$$
where:
- $R$ is the universal gas constant (8.31 J mol⁻¹ K⁻¹)
- $T_{b}$ is the boiling point of the solvent (in K)
- $\Delta H_{vap}$ is the enthalpy of vaporization (in J/mol)
According to the problem, the ratio of boiling points for X and Y is 2:1, and the ratio of their enthalpy of vaporization is 1:2.
Let's denote the boiling point of Y as $T_{b}(Y)$ and the boiling point of X as $T_{b}(X)$, and likewise for the enthalpy of vaporization $\Delta H_{vap}(Y)$ and $\Delta H_{vap}(X)$.
We then have:
$T_{b}(X) = 2T_{b}(Y)$ and $\Delta H_{vap}(X) = 0.5\Delta H_{vap}(Y)$
We can substitute these values into the equation for $K_{b}$:
$K_{b}(X) = R*(2T_{b}(Y))^2 / (1000 * 0.5\Delta H_{vap}(Y)) = 4*R*T_{b}(Y)^2 / (500 * \Delta H_{vap}(Y))$
and
$K_{b}(Y) = R*T_{b}(Y)^2 / (1000 * \Delta H_{vap}(Y))$
Comparing these two equations:
$K_{b}(X) / K_{b}(Y) = [4*R*T_{b}(Y)^2 / (500 * \Delta H_{vap}(Y))] / [R*T_{b}(Y)^2 / (1000 * \Delta H_{vap}(Y))] $
This simplifies to:
$K_{b}(X) / K_{b}(Y) = 4 / 0.5 = 8$
So, the boiling point elevation constant of X is 8 times the boiling point elevation constant of Y. Therefore, m = 8.
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