Question

The freezing point depression constant for water is 1.86$^{0}C$. If 5 g $N{{a}_{2}}S{{O}_{4}}$ is dissolved in 45 g of water, the freezing point is changed by -3.82$^{0}C$. Calculate the van’t Hoff factor for $N{{a}_{2}}S{{O}_{4}}$.
a.) 0.381
b.) 2.05
c.) 2.63
d.) 3.11

Answer 1

“The Van’t Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass”.

Complete step by step answer:
We know that the formula to calculate freezing point depression is $\Delta {{T}_{f}}={{K}_{f}}\times m\times i$
Where $\Delta {{T}_{f}}$= Depression in freezing point
${{K}_{f}}$=Freezing point constant
$m$= Molality of the solution
$i$= Van’t Hoff factor
In the question it is given that the mass of the solute i.e. $N{{a}_{2}}S{{O}_{4}}$= 5 g
We know that the Molar mass of solute = 142 g
Given mass of the solvent (water) = 45 g
$N{{a}_{2}}S{{O}_{4}}$= sodium sulfate

Now, by using above data we have to calculate the molality of the solution using the following formula.
m$=\dfrac{\text{mass of solute }\times \text{1000}}{\text{molar mass of solute }\times \text{mass of solvent in g}}$

& =\dfrac{5\times 1000}{142\times 45} \\\ & =0.78\text{ m} \\\ \end{aligned}$$ Molality of the solution = 0.78 m So, now we have $$\Delta {{T}_{f}}$$= -3.82 $$^{0}C$$ $${{K}_{f}}$$= - 1.86 $$^{0}C/m$$ $$m$$= 0.78 m $$i$$= ? By substituting all the above values in $$\Delta {{T}_{f}}={{K}_{f}}\times m\times i$$ we can calculate the value of Van't hoff's factor. $$\begin{aligned} & -{{3.82}^{0}}C=-1.86\dfrac{^{0}C}{m}\times 0.78\text{ }m\text{ }\times \text{ }i \\\ & i=\dfrac{3.82}{1.86\times 0.78} \\\ & i=2.63 \\\ \end{aligned}$$ Therefore the van’t Hoff factor for $$N{{a}_{2}}S{{O}_{4}}$$ is 2.63. **So, the correct answer is “Option C”.** **Additional Information:** “Van't Hoff factor is a measure of the effect of a solute on colligative properties like osmotic pressure, relative lowering in vapor pressure, elevation of boiling point and freezing point depression”. As the solute concentration increases the van't Hoff factor value decreases. **Note:** Don’t be confused there is no relation between concentrations of the solute to the freezing point depression of the solution. Because freezing point depression is directly proportional to the number of ions produced by the solute not on the concentration of the solute.