Question

A solution of sucrose (molar mass = 342 g / mol) has been prepared by dissolving 68.5 g of sucrose in 1000 g of water. The freezing point of the solution obtained will be: (\[{{K}_{f}}\]for water is \[1.86\text{ K kg / mol}\] )
(a)- \[-{{0.570}^{\circ }}C\]
(b)- \[-{{0.372}^{\circ }}C\]
(c)- \[-{{0.520}^{\circ }}C\]
(d)- \[+{{0.372}^{\circ }}C\]

Answer 1

Hint:
The depression in freezing point is calculated with the formula \[\Delta {{T}_{f}}={{K}_{f}}\text{ x }\dfrac{{{w}_{2}}}{{{M}_{2}}}\text{ x }\dfrac{1000}{{{w}_{1}}}\] , \[\Delta {{T}_{f}}\]is the depression in freezing point, \[{{w}_{2}}\] is the mass of the solute in gram, \[{{M}_{2}}\] is the molar mass of the solute and \[{{w}_{1}}\] is mass of the solvent in gram.

Complete step by step answer:
Let us understand about the depression in freezing point:
The freezing point of a substance is the temperature at which the solid and the liquid form of the substance are in equilibrium i.e., the solid and liquid form has the same vapor pressure. IIt is seen that the freezing point of the solution is always below the freezing point of the pure solvent. This is called the depression in freezing point.
The depression of freezing point depends on the solute in a solution and has found to be related to the molality’ as below:
\[\Delta {{T}_{f}}={{K}_{f}}\text{ x }m\]
\[{{K}_{f}}\]is the molal depression constant.
So, by expanding the above formula, we get:
\[\Delta {{T}_{f}}={{K}_{f}}\text{ x }\dfrac{{{w}_{2}}}{{{M}_{2}}}\text{ x }\dfrac{1000}{{{w}_{1}}}\]
Where, \[\Delta {{T}_{f}}\]is the depression in freezing point,
\[{{w}_{2}}\] is the mass of the solute in gram,
\[{{M}_{2}}\] is the molar mass of the solute,
\[{{w}_{1}}\] is the mass of the solvent in grams.
So, according to the question:
  \[{{w}_{2}}\] is the mass of solute = 68.5 g
  \[{{M}_{2}}\]= Molar mass of the solute = 342 g / mol
\[{{w}_{1}}\]= Mass of the solvent = 1000 g
Now, putting all these in the equation, we get
\[\Delta {{T}_{f}}={{K}_{f}}\text{ x }\dfrac{{{w}_{2}}}{{{M}_{2}}}\text{ x }\dfrac{1000}{{{w}_{1}}}=1.86\text{ x }\dfrac{68.5}{342}\text{ x }\dfrac{1000}{1000}=0.372\]
For calculating the freezing point of the solution we have to subtract the depression in freezing from the freezing point of the pure solvent.
\[\begin{align}
  & \Delta {{T}_{f}}={{T}^{\circ }}-{{T}_{f}} \\
 & {{T}_{f}}={{T}^{\circ }}-\Delta {{T}_{f}}=0-0.372=-{{0.372}^{\circ }}C \\
\end{align}\]
Since we know that the freezing point of pure water is 0.

Hence the correct option is (b)- \[-{{0.372}^{\circ }}C\].

Note:
\[{{K}_{f}}\] is also called the cryoscopic constant of the solvent. Depression in freezing point is a colligative property because it depends on the number of moles of the solute.