Question

For ${{0}}{{.6 ml}}$ of glacial acetic acid with density ${{1}}{{.06 gm }}{{{L}}^{{{ - 1}}}}$ is dissolved in ${{1kg}}$ water and solution froze at ${{ - 0}}{{.0205^\circ C}}$. Calculate van’t hoff factor and ${{{K}}_{{f}}}$ for water is ${{1}}{{.86 K Kg mo}}{{{l}}^{{{ - 1}}}}$ ?
(A)${{2}}{{.103}}$
(B)${{2}}{{.36}}$
(C)${{1}}{{.041}}$
(D)None of these.

Answer 1

Density is the ratio of mass of solution and volume of solution. By introducing the density and volume we can find the mass of solute easily. Then after finding moles from the mass of the solute we find the depression in freezing point.

Complete step by step answer:
From the given question above we collected the data that,
${\text{Volume of solute }}$ ${{V = 0}}{{.6 ml}}$
${\text{Density of solute}}$ $\rho {{ = 1}}{{.06 g m}}{{{l}}^{{{ - 1}}}}$
Mass of solute, ${{m}}$ can be calculated by multiplying density with volume of solute.
i.e. ${{m = }}\rho {{ \times V}}$
${{ = 0}}{{.6ml \times 1}}{{.06 g m}}{{{l}}^{{{ - 1}}}}$
${{ = 0}}{{.636 g}}$
Molecular mass of solute, ${{{M}}_{{s}}}{{ = 60 g mo}}{{{l}}^{{{ - 1}}}}$
Now we can calculate the number of moles, ${{n}}$ of solute by dividing mass of solute by molecular mass of solute.
i.e. ${{n = }}\dfrac{{{m}}}{{{{{M}}_{{s}}}}}$
By substituting the values, we get
{{n = }}\dfrac{{{{0}}{{.636 g}}}}{{{{60 g mo}}{{{l}}^{{{ - 1}}}}}}$$${{ = 0}}{{.0106 mol}}$ Now we have to calculate the molality. As we know that molality, ${{M}}$ can be calculated by dividing the number of moles by the mass of the solvent, ${{{m}}_{{s}}}$ in kilograms. ${{M = }}\dfrac{{{n}}}{{{{{m}}_{{s}}}}}$ Substituting the values, we get = \dfrac{{0.0106}}{1}$$$$= 0.0106$$
Depression in freezing point, ${{{T}}_{{f}}}$ can be calculated by multiplying depression constant, ${{{K}}_{{f}}}$ with molality.
i.e. ${{{T}}_{{f}}}{{ = }}{{{K}}_{{f}}}{{ \times m}}$
Substituting the values, we get
${{ = 1}}{{.86 K Kg mo}}{{{l}}^{{{ - 1}}}}{{ \times 0}}{{.0106 Kg mo}}{{{l}}^{{{ - 1}}}}$
${{ = 0}}{{.01971 K}}$
Van’t Hoff factor is obtained by dividing the observed depression in freezing point by normal depression in freezing point.
i.e. ${{ = }}\dfrac{{{{0}}{{.0205K}}}}{{{{0}}{{.01971K}}}}{{ = 1}}{{.0400}}$
Therefore, the correct option is (3).

Additional Information:
Van't Hoff factor is a calculative value of the effect of a solute on various colligative properties of the solution like relative lowering of the vapor pressure, the osmotic pressure, depression in the freezing point and the elevation in the boiling point of the solution.
Molality is the property of a solution and it defines the number of moles of solute per kilogram of the solvent.
We should be careful with the molarity and the molality a we might get confused between the two
Molarity is the number of moles in a liter of solution while molality is the number of moles in a kilogram of solvent. The units are of huge difference

Note:
Special attention should be paid while converting the volume of the solution in mass. The units of density should be converted in the units of volume itself to avoid further difference in answer.