Question

Mercury(II) chloride (HgCl₂)freezes at$\mathbf{276}\mathbf{.}\mathbf{1}\mathbf{°}\mathbf{C}$and has a freezing-point depression constant K_fof 34.3 Kkgmol^-1. When 1.36gof solid mercury(I) chloride (empirical formula HgCl) is dissolved in 100gof (HgCl₂), the freezing point is reduced by $\mathbf{0}\mathbf{.}\mathbf{99}\mathbf{°}\mathbf{C}$. Calculate the molar mass of the dissolved solute species and give its molecular formula.

Short answer

The molar mass of the solute is 471.2g/mol and the molecular formula is Hg₂Cl₂ .

Step-by-step solution

Given data

The given information are given as:

Mass of solute = 1.36g.

Mass of solvent = 100g.

Depression in freezing point = 0.99K.

Depression in freezing point constant =34.3K.kg/mol.

Concept of Colligative properties

The properties which depend on the number of moles of solute irrespective of its nature are termed as colligative properties. Relative lowering in vapor pressure, depression in freezing point, and osmotic pressure are examples of colligative properties.

Derive an expression for molality

The expression for depression in freezing point is represented as follows:

$∆{\mathrm{T}}_{\mathrm{f}}={\mathrm{K}}_{\mathrm{f}}.\mathrm{m}$ ..........(1)

Where,

m is molality.

K_fis depression in freezing point constant.

$∆{\mathrm{T}}_{\mathrm{f}}$is depression in freezing point.

Molality is defined as the number of moles of solute dissolved in 1 kg of the solvent. The expression for molality is represented as follows:

$$\begin{gathered} \mathrm{Molality}=\frac{\mathrm{Number}\mathrm{of}\mathrm{moles}\mathrm{of}\mathrm{solute}}{\mathrm{Mass}\mathrm{of}\mathrm{solvent}\mathrm{in}\mathrm{kg}} \\ \mathrm{m}=\frac{\mathrm{Given}\mathrm{mass}}{\mathrm{Molar}\mathrm{mass}\times \mathrm{Mass}\mathrm{of}\mathrm{solvent}\mathrm{in}\mathrm{kg}} \end{gathered}$$

Calculation for the molar mass of solute

Plug the given values in equation 1.

$$\begin{gathered} ∆{\mathrm{T}}_{\mathrm{f}}=\frac{{\mathrm{K}}_{\mathrm{f}}\times \mathrm{Givenmass}}{\mathrm{Molar}\mathrm{mass}\mathrm{of}\mathrm{solute}\times \mathrm{Mass}\mathrm{of}\mathrm{solvent}\mathrm{in}\mathrm{kg}} \\ \mathrm{Therefore},\mathrm{the}\mathrm{molar}\mathrm{mass}\mathrm{of}\mathrm{solute}\mathrm{is}\mathrm{given}\mathrm{as}: \\ \mathrm{Molar}\mathrm{mass}\mathrm{of}\mathrm{solute}=\frac{{\mathrm{K}}_{\mathrm{l}}\times \mathrm{Givenmass}}{\mathrm{Mass}\mathrm{of}\mathrm{solvent}\mathrm{in}\mathrm{kg}\times ∆{\mathrm{T}}_{\mathrm{f}}} \\ \mathrm{Molar}\mathrm{mass}\mathrm{of}\mathrm{solute}=\frac{{\mathrm{K}}_{\mathrm{f}}\times \mathrm{Givenmass}}{\mathrm{Mass}\mathrm{of}\mathrm{solvent}\mathrm{in}\mathrm{kg}\times ({\mathrm{T}}_0-{\mathrm{T}}_{\mathrm{f}})} \\ \mathrm{Molar}\mathrm{mass}\mathrm{of}\mathrm{solute}=\frac{34.3\mathrm{K}.\mathrm{kg}/\mathrm{mol}\times 1.36\mathrm{g}}{100\mathrm{g}\times {\displaystyle \frac{1\mathrm{kg}}{{1000}_{\mathrm{g}}}}\times (0.99)\mathrm{K}} \\ \mathrm{Molar}\mathrm{mass}\mathrm{of}\mathrm{solute}=\frac{34.3\mathrm{K}.\mathrm{kg}/\mathrm{mol}\times 1.36\mathrm{g}}{0.1\mathrm{kg}\times (0.99)\mathrm{K}} \\ \mathrm{Thus},\mathrm{the}\mathrm{molar}\mathrm{mass}\mathrm{of}\mathrm{the}\mathrm{solute}\mathrm{is}471.2\mathrm{g}/\mathrm{mol}. \end{gathered}$$

Calculation for the molecular formula

Now, the molecular formula will be calculated as follows:

$$\begin{gathered} \mathrm{n}=\frac{\mathrm{Molarmass}}{\mathrm{Empirical}\mathrm{formulamass}} \\ \mathrm{n}=\frac{471.2\mathrm{g}/\mathrm{mol}}{\mathrm{HgCl}} \\ \mathrm{n}=\frac{471.2\mathrm{g}/\mathrm{mol}}{271.52\mathrm{g}/\mathrm{mol}}=1.7 \\ \mathrm{n}=2 \end{gathered}$$

Since, molecular formula is given as:

$$\begin{gathered} \mathrm{Molecular}\mathrm{formula}=\mathrm{n}\times \mathrm{Empirical}\mathrm{formula} \\ \mathrm{Molecular}\mathrm{formula}=2\times \mathrm{HgCl} \\ \mathrm{Molecular}\mathrm{formula}={\mathrm{Hg}}_2{\mathrm{Cl}}_2 \end{gathered}$$

Therefore, the molecular formula for the compound is Hg₂Cl₂.