Question

(a) Define the following terms
(i) Mole fraction
(ii) Ideal solution
(b) $15.0{\text{ g}}$ of an unknown molecular material was dissolved in $450{\text{ g}}$ of water. The resulting solution was found to freeze at $- {0.34^{\text{o}}}{\text{C}}$ . What is the molar mass of this material? ( ${K_f}$ for water $= 1.86{\text{ K kg mol}}$ )

Answer 1

(a) (i) Mole fraction deals with the number of moles of solute and solvent. It is a method used to express the concentration of solute in a solution.
(ii) Solutions can be classified into two types, ideal solution and non-ideal solution, depending on their behavior. Non-ideal solutions show positive and negative deviations from Raoult’s law.
(b) Use the following formula for the depression in the freezing point to calculate the molar mass of solute.
$\Delta {T_f} = \dfrac{{1000 \times {K_f} \times {W_2}}}{{{W_1} \times {M_2}}}$
Here, $\Delta {T_f}$ represents the depression in the freezing point, ${K_f}$ represents the molal depression in the freezing point constant, ${W_2}$ represents the mass of the solute, ${M_2}$ represents the molar mass of the solute and ${W_1}$ represents the mass of the solvent.

Complete answer:
(a) Define the following terms
(i) Mole fraction of a component in a mixture of two or more components, is the ratio of the number of moles of that component to the total number of moles of all the components in the solution. For example, consider a binary solution containing solute A and solvent A. Let ${n_A}$ and ${n_B}$ represent the number of moles of solute A and solvent B respectively. Also, let ${\chi _A}$ and ${\chi _B}$ represent the mole fractions of solute A and solvent B respectively. Then

$$\Rightarrow {\chi _A} = \dfrac{{{n_A}}}{{{n_A} + {n_B}}} \\\ \Rightarrow {\chi _B} = \dfrac{{{n_B}}}{{{n_A} + {n_B}}} \\\ \Rightarrow {\chi _A} + {\chi _B} = 1 \\\$$

In a binary solution, the sum of the mole fractions of solute and solvent is equal to unity.

(ii) Ideal solution obeys Raoult’s law over a wide range of temperature and pressures. For a binary ideal solution, the total vapour pressure of the solution is equal to that predicted by Raoult's law.

$${P_T} = {P_A} + {P_B} \\\ {P_T} = \left( {P_A^o \times {\chi _A}} \right) + \left( {P_B^o \times {\chi _B}} \right) \\\$$

(b)Use the following formula for the depression in the freezing point to calculate the molar mass of solute.
$\Rightarrow \Delta {T_f} = \dfrac{{1000 \times {K_f} \times {W_2}}}{{{W_1} \times {M_2}}}$
Rearrange the above formula
$\Rightarrow {M_2} = \dfrac{{1000 \times {K_f} \times {W_2}}}{{{W_1} \times \Delta {T_f}}}$
Substitute values and calculate the molar mass of solute

$$\Rightarrow {M_2} = \dfrac{{1000 \times {K_f} \times {W_2}}}{{{W_1} \times \Delta {T_f}}} \\\ \Rightarrow {M_2} = \dfrac{{1000 \times 1.86 \times 15}}{{450 \times 0.34}} \\\ \Rightarrow {M_2} = 182.35{\text{ g/mol}} \\\$$

Hence, the molar mass of solute is $182.35{\text{ g/mol}}$ .

Note:
(a) A binary solution contains only two components. The component present in small amounts is the solute and the component present in large amounts is solvent.
(b)The depression in the freezing point is the colligative property. The depression in the freezing point depends only on the number of solute particles and is independent of the identity of the solute.