Question

How many moles of sucrose should be dissolved in 500 gms of water so as to get a solution which has a difference of 104${}^\circ $C between boiling point and freezing point? (${{K}_{f}}=1.86Kkgmo{{l}^{-1}}$ , ${{K}_{b}}=0.52Kkgmo{{l}^{-1}}$)
(A) 1.68
(B) 3.36
(C) 8.40
(D) 0.840

Answer 1

Hint: The concepts for solving the question is based on two colligative properties i.e. boiling point elevation and depression in freezing point.

Complete step by step answer:
Let’s start with understanding the concept of depression in freezing point (Cryoscopy) and elevation is boiling point (Ebullioscopy).

Freezing point-
The temperature at which the vapour pressure of liquid is equal to the vapour pressure of solid formed is called the freezing point.
Vapour pressure of liquid = Vapour pressure of solid

Depression in freezing point of the solution-
On adding non-volatile solute to pure solvent, the freezing point of the solution will decrease than that of pure solvent and this decrease is called depression in freezing point. The liquid solution has a lower freezing point than pure solvent because the chemical potential of the solvent in mixture is lower than that of -pure solvent.
$\Delta {{T}_{f}}={{T}_{f}}{}^\circ -{{T}_{f}}$
where,
$\Delta {{T}_{f}}$ = freezing point depression
${{T}_{f}}{}^\circ $= freezing point of pure solvent
${{T}_{f}}$ = freezing point of solution
For dilute solutions,
$\begin{align}
 & \Delta {{T}_{f}}\propto m \\
 & \Delta {{T}_{f}}={{K}_{f}}\times m\times i \\
\end{align}$
where,
m = molality
i = Vant Hoff factor
${{K}_{f}}$ = cryoscopic constant.

Boiling point-
The temperature at which the vapour pressure of liquid equals the vapour pressure of the environment is called the boiling point.
Vapour Pressure of liquid = Vapour Pressure of environment
Elevation in boiling point-
On adding the non-volatile solute to a solvent, the resulting solution will have higher boiling point than that of pure solvent. This rise in boiling point is known as elevation in boiling point.
\[\Delta {{T}_{b}}={{T}_{b}}-{{T}_{b}}{}^\circ \]
where,
\[\Delta {{T}_{b}}\] = boiling point elevation
\[{{T}_{b}}\] = boiling point of solution
\[{{T}_{b}}{}^\circ \] = boiling point of pure solvent

For dilute solutions,
$\begin{align}
 & \Delta {{T}_{b}}\propto m \\
 & \Delta {{T}_{b}}={{K}_{b}}\times m\times i \\
\end{align}$
where,
m = molality
i = Vant hoff factor
${{K}_{b}}$ = ebullioscopic constant

Now,
Let us see the given data in the illustration mentioned:
Mass of solvent = 500 gms
${{T}_{b}}-{{T}_{f}}=104{}^\circ C$
${{K}_{f}}=1.86Kkgmo{{l}^{-1}}$
${{K}_{b}}=0.52Kkgmo{{l}^{-1}}$
As we know,
Boiling point of water, ${{T}_{b}}{}^\circ $= 100${}^\circ C$
Freezing point of water, ${{T}_{f}}{}^\circ =0{}^\circ C$

Thus, using above mentioned equations,
$\Delta {{T}_{b}}+\Delta {{T}_{f}}=\left( {{T}_{b}}-{{T}_{f}} \right)+\left( {{T}_{f}}{}^\circ -{{T}_{b}}{}^\circ \right)$
$\Delta {{T}_{b}}+\Delta {{T}_{f}}$ = (104) + (0-100)
$\Delta {{T}_{b}}+\Delta {{T}_{f}}$ = 4${}^\circ C$
Now, using equations of $\Delta {{T}_{b}}$and $\Delta {{T}_{f}}$;
$\Delta {{T}_{b}}+\Delta {{T}_{f}}=\left( {{K}_{b}}\times m\times i \right)+\left( {{K}_{f}}\times m\times i \right)=4{}^\circ C$

As, sucrose is a nonelectrolyte, i = 1
$m\left( {{K}_{b}}+{{K}_{f}} \right)=4{}^\circ C$
$m=\dfrac{4{}^\circ C}{\left( 0.52+1.86 \right){}^\circ CKg/mol}$
m = 1.68m
Since, we have 500 gm of solvent.
The moles of water in water can be calculated as,
Moles of sucrose = $1.68m\times 0.500Kg$
 = 0.840 moles.
So, the correct answer is “Option D”.

Note: Do note that we should use the proper and standard units while calculating the temperatures and also molality.
As sucrose never dissociates, we took Van't Hoff factor as 1.