Question

A solution containing $28\;{\text{g}}$ of phosphorus in $315\;{\text{g}}$ of ${\text{C}}{{\text{S}}_2}$ (boiling point $\left. {{{46.3}^\circ }{\text{C}}} \right)$ boils at ${47.98^\circ }{\text{C}}$. The molal elevation constant for ${\text{C}}{{\text{S}}_2}$ is $2.34\;{\text{K}}\;{\text{kg}}\;{\text{mo}}{{\text{l}}^{ - 1}}$. Calculate the molecular mass of phosphorus and deduce its formula in ${\text{C}}{{\text{S}}_2}$.

Answer 1

Hint: A quality of a solution known as a colligative property depends on the ratio of the total number of solute particles to the total number of solvent particles in the solution. The chemical makeup of the solution's constituents has no bearing on the colligative properties. We use this concept here to solve the problem.

Complete Step by Step Solution:
Colligative qualities are those of a solution that depend on the proportion of solute to solvent particles present, rather than the makeup of the chemical species, in the solution.
When a solute is added, the boiling point of a solvent rises, a process known as boiling point elevation. The solution formed when a non-volatile solute is dissolved in a solvent has a higher boiling point than the solvent alone. For instance, a solution of sodium chloride (salt) and water has a boiling point higher than pure water.

The solute-to-solvent ratio, but not the solute's identity, determines the boiling point elevation, which is a collative feature of matter. This suggests that the amount of solute added to a solution affects how high its boiling point rises. The boiling point will rise proportionately to the solute concentration in the solution.

The following is an expression for the boiling point of a solution containing a non-volatile solute: Boiling point of a solution is calculated by adding the boiling point of a pure solvent.
The concentration of the solute in the solution is inversely correlated with the increase in boiling point. It can be calculated using the equation below:
$\Delta {T_b} = m{K_b}$

The following formula is used
$ \Delta {T_b} = m{K_b} = \frac{w}{M} \times \frac{{1000}}{W} \times {K_b} \\$
$ \Delta {T_b} = 47.98 - 46.3 \\$
$ = 1.681.68 = \frac{{28}}{M} \times \frac{{1000}}{{315}} \times 2.38 \\$
$ M = \frac{{28 \times 1000 \times 2.38}}{{315 \times 1.68}} = 125.92 \\$
$ {\text{Atomicity }} = \frac{{{\text{ Mol}}{\text{.wt}}{\text{. }}}}{{{\text{ At}}{\text{.wt}}{\text{. }}}} = \frac{{125.92}}{{31}} = 4.02 \\ $

So the Molecule is $ = {P_4}$.

Note: The total number of atoms in a molecule is known as its atomicity. For instance, two oxygen atoms make up each oxygen molecule. So, oxygen has an atomic number of 2. Atomicity is occasionally used in the same sense as valency in earlier settings.