Question

The molal freezing point constant for water is 1.86. If 342 g of cane sugar is dissolved in 1000 g of water, the solution will freeze at:
A. ${{1.86}^{\text{o}}}\text{C}$
B. $-{{1.86}^{\text{o}}}\text{C}$
C. $-{{3.92}^{\text{o}}}\text{C}$
D. ${{3.92}^{\text{o}}}\text{C}$

Answer 1

Hint: When any substance or solute is added to any solvent, it experiences the decrement in its freezing point. Depression in freezing point is actually the decrement in freezing point of a solvent on the addition of a non-volatile solute. Use the formula $\vartriangle {{\text{T}}_{\text{f}}}={{\text{K}}_{\text{f}}}\times \text{m}\times \text{i}$, to find the depression in freezing point.

Complete step by step answer:
Considering as an ideal solution, the depression depends on the solute concentration that can be expressed as $\vartriangle {{\text{T}}_{\text{f}}}={{\text{K}}_{\text{f}}}\times \text{m}\times \text{i}$; where
- $\vartriangle {{\text{T}}_{\text{f}}}$ is the freezing-point depression expressed as ${{\text{T}}_{\text{f}}}$ (pure solvent) − ${{\text{T}}_{\text{f}}}$ (solution).
- ${{\text{K}}_{\text{f}}}$ is the cryoscopic constant which only depends on the properties of the solvent.
- $\text{m}$ is the molality which is represented as moles solute per kilogram of solvent or $\dfrac{\text{moles of solute}}{\text{mass of solvent in kg}}$.
- $\text{i}$ is van't Hoff factor (number of ions per molecule of solute. i = 2 for $\text{NaCl}$ and for non-electrolytes, i=1 like cane sugar).
Now, find molality by finding the moles of cane sugar; moles are defined as the weight of substance per molar mass of it.
Its formula is $\text{moles=}\dfrac{\text{mass of substance}}{\text{molar mass}}$. Molar mass of cane sugar $\left( {{\text{C}}_{12}}{{\text{H}}_{22}}{{\text{O}}_{11}} \right)$:
- Atomic mass of carbon is 12 grams.
- Atomic mass of hydrogen is 1 gram.
- Atomic mass of oxygen is 16 grams.
Molar Mass is $\left[ \left( \text{12}\times \text{12} \right)+\left( 22\times 1 \right)+\left( 16\times 11 \right) \right]$ or 342 grams.
Let us solve the numerical, first of all, write the values given in the question.
Mass of cane sugar = 342 grams and mass of water (solvent) = 1000g, i=1 (non-electrolyte) and ${{\text{K}}_{\text{f}}}$ = 1.86 K kg$\text{mo}{{\text{l}}^{-1}}$.
The molality becomes $\dfrac{\text{342}\times \text{1000}}{342\times 1000}$. The term is divided by 1000 because we have to find per ‘kg’ of solvent. The value of molality is 1 m.
Now, put the value of m in the formula $\vartriangle {{\text{T}}_{\text{f}}}={{\text{K}}_{\text{f}}}\times \text{m}\times \text{i}$; so, $\vartriangle {{\text{T}}_{\text{f}}}=1.86\times 1\times 1$ , $\vartriangle {{\text{T}}_{\text{f}}}$ will be 1.86$^{\text{o}}\text{C}$.
The freezing point of solution will be ${{\text{T}}_{\text{f}}}$ (pure solvent) − ${{\text{T}}_{\text{f}}}$ (solution).
The solvent here is water; ${{\text{T}}_{\text{f}}}$= ${{0}^{\text{o}}}\text{C}$ and $\vartriangle {{\text{T}}_{\text{f}}}={{1.86}^{\text{o}}}\text{C}$ , so, ${{\text{T}}_{\text{f}}}$(solution) will be ${{0}^{\text{o}}}\text{C}-{{1.86}^{\text{o}}}\text{C = }-{{1.86}^{\text{o}}}\text{C}$.

The answer to this question is option ‘b’, the freezing point of the solution is -1.86$^{\text{o}}\text{C}$.

Note: The common mistake that students commit here is not using the formula of molality correctly. Do not forget to divide by ‘1000’ as the formula is moles of solute per kg of solvent. So, remember to convert the mass of solvent which may be given in grams to convert that in kilograms. Depression in freezing point is a colligative property. It depends on the number of solute particles. Colligative properties are directly proportional to number of solute particles.