Question

A solution of x moles of sucrose in ${{100g}}$ of water freezes at ${{ - 0}}{{.2^\circ C}}$. As ice separates the freezing point goes down to ${{0}}{{.25^\circ C}}$. How many grams of ice would have separated?
(A) ${{18g}}$
(B) ${{20g}}$
(C) ${{25g}}$
(D) ${{23g}}$

Answer 1

Hint: First we should know about the colligative properties of the solution and why the freezing point, boiling point, vapour pressure and osmotic pressure changes and then we will find out the amount converted to ice.

Complete step by step answer:
When we mix a non – volatile solute in a solution some properties of the solution changes because of surface area being occupied by the non – volatile solute so less surface area is left for the solvent to vaporize. Due to these four properties known as colligative properties are changed drastically namely the vapour pressure, boiling point, freezing point, osmotic pressure. So, after mixing of solute there is relative lowering in vapour pressure, elevation in boiling point, depression in freezing point and osmotic pressure.
The actual freezing point of the pure solvent is greater than the freezing point of the solution after mixing solute.
${{\Delta }}{{{T}}_{{f}}}{{ = T}}_{{f}}^{{o}}{{ - }}{{{T}}_{{f}}}$, where $\Delta {{{T}}_{{f}}}$ is the depression in freezing point, ${{T}}_{{f}}^ \circ $ is the actual freezing point and ${{{T}}_{{f}}}$ is the freezing point of solution.
This depression in freezing point is directly dependent on molality of the solution, ${{m}}$.
i.e. ${{\Delta }}{{{T}}_{{f}}} \propto {{m}}$
${{\Delta }}{{{T}}_{{f}}}{{ = }}{{{K}}_{{f}}}{{ \times m}}$, where ${{{K}}_{{f}}}$ is the freezing point depression constant
Here ${{{T}}_{{f}}}{{ = - 0}}{{.2^\circ C T}}_{{f}}^{{o}}{{ = 0^\circ C}}$ for water molecule.
${{\Delta }}{{{T}}_{{f}}}{{ = 0}}{{.2 - 0 = 0}}{{.}}{{{2}}^ \circ }{{C}}$
It is given that there are ${{x}}$ moles of sucrose in $100{{g}}$ of water.
${{\Delta }}{{{T}}_{{f}}}{{ = }}{{{K}}_{{f}}}{{ \times m}}$
Molality can be expressed as ${{m = }}\dfrac{{{{xmol}}}}{{100 \times {{10}^{ - 3}}{{kg}}}}$
Substituting the value of molality in the above equation, we get
${{\Delta }}{{{T}}_{{f}}}{{ = }}{{{K}}_{{f}}}{{ \times }}\dfrac{{{{xmol}}}}{{100 \times {{10}^{ - 3}}{{kg}}}}$
Putting value of depression constant, we get
${{0}}{{.2 = }}\dfrac{{{{1}}{{.86 \times x \times 1000}}}}{{{{100}}}}$
On solving, we get
${{x = 0}}{{.01075 mol}}$
Now assume that ${{w}}$ amount of ice freezes so remaining water will be ${{100 - w}}$
Now we have ${{{T}}_{{f}}}{{ = - 0}}{{.25^\circ C}}$
${{\Delta }}{{{T}}_{{f}}}{{ = }}\dfrac{{{{{K}}_{{f}}}{{ \times x \times 1000}}}}{{{{(100 - w)}}}} \\
{{0}}{{.25 = }}\dfrac{{{{1}}{{.86 \times 0}}{{.01075 \times 1000}}}}{{{{(100 - w)}}}} $
${{w = 20 g}}$

So, the correct answer is Option B.

Note: When calculating molality or molarity special attention should be paid on the unit of volume or mass of solvent. If the unit is in ml or grams then it should be divided by one thousand to convert it into litre or Kilograms respectively.
The depression in freezing point occurs because of lowering in vapour pressure. The vapour pressure is decreased because of reduced surface area of solvent molecules which are occupied by solute molecules and less vaporization happening.