Question

$1$ m aq. Solution of a salt shows elevation in boiling point equal to $3$ times of its ${K_b}$ . Determine:
-Freezing point of solution
-Osmotic pressure at ${27^ \circ }C$
${K_f}\left( {water} \right) = 1.86\;K\;Kg\;mol{e^{ - 1}}$

Answer 1

Boiling point elevation is that the development within which the boiling point of the liquid would be higher when another molecule is added, which implies that the solution’s boiling point is outrageous than that of a pure solvent.

Complete answer:
The raising of the solvent’s boiling point is understood as boiling point elevation that is due to the addition of the solute. Once the boiling point will increase then the freezing point decreases.
The formula of boiling point elevation is:
$\Delta {T_b} = {K_b} \times {m_B}$
Where,
$\Delta {T_b}$ is that the boiling point elevation that is outlined as ${T_{b\left( {solution} \right)}} - {T_{b\left( {pure\;solvent} \right)}}$ .
${K_b}$ is the ebullioscopic constant whose value is dependent on the properties of the solvent. It may be calculated as ${K_b} = \dfrac{{RT_b^2M}}{{\Delta {H_v}}}$ , where $R$ is that the universal gas constant and ${T_b}$ is that the boiling temperature of the pure solvent, $M$ is that the molar mass of the solvent and heat of vaporization per mole of the solvent is $\Delta {H_v}$ .
Here, the molality of the solution is ${m_B}$ which might be calculated by taking dissociation into consideration as the boiling point elevation is considered a colligative property that depends on the quantity of particles in solution. This is often most simply done by using the Van’t Hoff factor $i$ as
${m_B} = {m_{solute}} \times i$ .
The factor $i$ accounts for the quantity of individual particles formed by a compound in solution.
$\Delta {T_b} = i \times {K_b} \times m$
$\Rightarrow 3{K_b} = i \times {K_b} \times 1$
$\Rightarrow i = 3$
The temperature within which the liquid solvent and therefore the solid solvent are at equilibrium, so their vapour pressures are equal is termed as the freezing point. On the addition of non-volatile solute and a volatile liquid solvent, the solution vapour pressure is not up to that of the pure solvent. The freezing point within the given question is:
$\Delta {T_f} = i \times {K_f} \times m$
$= 3 \times 1.86 \times 1$
Or $\Delta {T_f} = - {5.58^ \circ }C$
The minimum pressure that is needed to be applied to a solution to stop the inward flow of its pure solvent across the membrane is termed as osmotic pressure. The osmotic pressure within the given question is:
$\pi = iCRT$
$\Rightarrow 3 \times 1 \times 0.0821 \times 300$
$\Rightarrow 73.89\;atm$ .

Note:
The development of boiling point elevation happens where a non-volatile solute like salt is added to a pure solvent like water. The change in the boiling point is more than the change in freezing point at different altitudes.