Question

What mass of sugar $${\mathrm{C}}_{12}{\mathrm{H}}_{22}{\mathrm{O}}_{11}$$ ($${\mathrm{M}}_0=342$$) must be dissolved in $$4.0\mathrm{k}\mathrm{g}$$of$${\mathrm{H}}_2\mathrm{O}$$ to yield a solution that will freeze at $$-3.72{}^0\mathrm{C}$$?
(Take $$K_{\mathrm{f}}=1.86{}^0\mathrm{C}{\mathrm{m}}^{-1}$$).

Answer 1

Hint:
As we know that, the colligative properties depend upon the number of solute particles. There are four types of colligative properties in which by one property, the above question can be solved.

Complete step by step solution
It is known that as the sugar (solute) is dissolved in water it occupies the position of water and now there is a small number of molecules of water so the freezing point becomes low and that is known by depression in freezing point.

For depression in freezing point of a solvent, we have a relationship between the depression in the freezing point and molality of the solution that can be written as follows:
$\mathrm{\Delta }{T}_f=K_f\mathrm{m}$

Here, $\mathrm{\Delta }{T}_f$ is represented as the depression in the freezing point of the solvent, $K_f$ is the molal depression constant for the given solvent and $\mathrm{m}$ is the molality of the solution.
We are given as
$$K_{\mathrm{f}}=1.86{}^0\mathrm{C}{\mathrm{m}}^{-1}=274.86\mathrm{K}{\mathrm{m}}^{-1}$$
$${\mathrm{M}}_0=342$$
$${\mathrm{M}}_2=4\mathrm{k}\mathrm{g}$$
$$\mathrm{\Delta }{\mathrm{T}}_{\mathrm{f}}$$=$$-3.72{}^0\mathrm{C}=269.28\mathrm{K}$$
Now, first of all we will calculate the molality by the above equation.
$$\begin{gathered} 269.28\mathrm{K}=274.86\mathrm{K}{\mathrm{m}}^{-1}\times \mathrm{m} \\ \mathrm{m}={\displaystyle \frac{269.28\mathrm{K}}{274.86\mathrm{K}{\mathrm{m}}^{-1}}} \\ =0.98\mathrm{m} \end{gathered}$$
Now, we know that molarity is defined as the number of solute particles dissolved in per kilogram of solvent. The solvent is water and suppose $${\mathrm{w}}_0$$ is weight of sugar which is our question.
$$molality(m)={\displaystyle \frac{({\mathrm{w}}_0/{\mathrm{M}}_0)}{\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{v}\mathrm{e}\mathrm{n}\mathrm{t}(\mathrm{g})}}\times 1000---(ii)$$
Now, putting the values in the above equation, we get,
$$\begin{gathered} {\mathrm{w}}_0=4\mathrm{k}\mathrm{g}\times 0.98\mathrm{m}\times 0.342{\displaystyle \frac{\mathrm{k}\mathrm{g}}{\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}}} \\ =1.34\mathrm{k}\mathrm{g} \end{gathered}$$

Therefore, the mass of the sugar is $$1.34\mathrm{k}\mathrm{g}$$.

Note:
We can also calculate the other values by using the above formula such as depression in freezing point, weight of solvent etc.