Question

If $10g of solid dissolved in$200g of water and the resulting solution has a freezing point $- {3.72^ \circ }c$ then determine the molar mass of non-ionizing solid?

A.$25g/mol$
B. $50g/mol$
C. $100g/mol$
D.$150g/mol$

Answer 1

In order to find the solution of this question, first we need to find the morality of the given solution. Then we have to equate it with the depression of the freezing point of the solid.

Formula used: $\Delta {T_f} = {K_f}m$ where, $\Delta {T_f}$is depressed in freezing point. ${K_f} =$Freezing point constant and $m =$molality.

Complete step-by-step solution:
It is provided that the mass of the water is $200g$or $0.2kg$.
We know that molality $m = \dfrac{n}{V}$where $n =$number of moles of solute and ${m_2} =$mass of solvent (in kg).
Again, molality $m = \dfrac{{{m_1}}}{{M{m_2}}}$where ${m_1} =$ mass of solute (in kg), $M =$molar mass of solute and ${m_2} =$mass of solvent (in kg).
We discussed earlier that:
$\Rightarrow \Delta {T_f} = {K_f}m - - (i)$
${m_1} = 10g$, ${m_2} = 0.2kg$, the molal depression in freezing point constant ${K_f} = 1.86K/m$. The freezing point of pure water is ${0^ \circ }c$and freezing point of solution is $- {3.72^ \circ }c$. Hence, depression in freezing point,
$\Delta {T_f} = {\left( {0 - ( - 3.72)} \right)^ \circ }c \\\ \Delta {T_f} = {3.72^ \circ }c \\\$
Substituting all these above values within equation$(i)$ we get:
$\Rightarrow {3.72^ \circ }c = 1.86 \times \dfrac{{10}}{{M \times 0.2}} \\\ \Rightarrow M = \dfrac{{1.8 \times 10}}{{3.72 \times 0.2}} \\\ \Rightarrow M = 25g/mol \\\$

The molar mass of non-ionizing solid is option$(a)$ $M = 25g/mol$.

Note: Colligative properties: Colligative properties of solutions are properties that depend upon the concentration of solute not upon the concentration of its ion.
Colligative properties can be classified into 4 types:
Lowering the vapour pressure: it can be illustrated mathematically as follows:
$P = {X_{solvent}}{p^ \circ }$ where ${X_{solvent}} =$mole fraction of solvent, ${P^ \circ } =$Vapour pressure of pure solvent
This law is also called Raoult’s law.

Boiling point elevation: It can be illustrated mathematically as follows:
$\Delta {{\rm T}_b} = {K_b}m$ where ${K_b} =$molal boiling point elevation constant or ebullioscopic constant.
Freezing point depression: It can be illustrated mathematically as follows:
$\Delta {T_f} = {K_b}m$ where ${K_f} =$molal freezing point depression constant or Cryoscopic constant.
Osmatic pressure$(\pi )$: It can be illustrated mathematically as follows:
$\pi = cRT$where $c =$concentration of solvent, $R =$universal gas constant, $T =$temperature.