Question

The molal freezing point constant of water is $1.86{\text{ K }}{{\text{m}}^{ - 1}}$. If $342{\text{g}}$ of cane sugar ${C_{12}}{H_{22}}{O_{11}}$ is dissolved in $1000{\text{g}}$ 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.${2.42^{\text{o}}}{\text{C}}$

Answer 1

Hint: To answer this question, you must recall the colligative properties, mainly, depression in freezing point. When a solvent or impurity is added in any solvent, it experiences a decrease in the freezing point. So, we can define the depression in freezing point as the decrease in the freezing point of a solvent experienced on the addition of a non- volatile impurity or solute.
Formula used:
$\Delta {T_f} = {K_f} \times m$
Where, $\Delta {T_f}$ represents the depression in freezing point of the mixture
${K_f}$ represents the molal freezing point constant or the cryoscopic constant.
And $m$ represents the molality of the solution which is given as,
${\text{m}} = \dfrac{n}{{W({\text{kg}})}}$
Where, $n$ is the number of moles of the solute present in the given solution
$W$ is the weight of the solvent given in kilograms

Complete step by step answer:
We are supposed to calculate the freezing point of the solution so formed after the given amount of cane sugar is added into water. We can find it using the depression in the freezing point, for which we first have to calculate the molality of the solution.
We know that the molar mass of cane sugar or sucrose can be calculated as $342{\text{g}}$
The given mass of cane sugar added in $1000{\text{g}}$ water is also given to be $342{\text{g}}$. We can write the molality of the solution as ${\text{m}} = \dfrac{n}{{W({\text{kg}})}}$
$ \Rightarrow m = \dfrac{{342}}{{342{\kern 1pt} \times 1000}} \times 1000$
$ \Rightarrow m = 1$
We know the expression for the depression in freezing point as $\Delta {T_f} = {K_f} \times m$
$\Delta {T_f} = 1.86 \times 1 \times 1 = 1.86{\text{ K or 1}}{\text{.8}}{{\text{6}}^{\text{o}}}{\text{C}}$
We know that the freezing point of water is ${0^{\text{o}}}{\text{C}}$. Thus, the freezing point of the solution after addition of cane sugar will be $T = 0 - 1.86 = - {1.86^{\text{o}}}{\text{C}}$

Thus, the correct answer is A.
Note:
Colligative properties are those properties of a solution which depend only on the number of the solute particles present in the solution. They are: elevation in boiling point, depression in freezing point, lowering of vapour pressure and osmotic pressure.