Question

12.2g of benzoic acid $({{M}_{w}}=122)$ in 100g benzene has depression in freezing point ${{2.6}^{0}}$; ${{K}_{f}}={{5.2}^{0}}kg/mol$. if there is 100% polymerisation, the number of molecules of benzoic acid in dissociated state is
a) 1
b) 2
c) 3
d) 4

Answer 1

Hint: The answer to this question is based on the formula which is used to calculate the depression in the freezing point of the solution which is given by $\Delta {{T}_{f}}=i{{K}_{f}}m$ and substitution of values gives the required answer.

Complete step – by – step solution:
The concepts of physical chemistry which deals with the calculation of various parameters that are included in the thermodynamics and also the colligative properties part are familiar to us.
Let us now calculate the number of molecules associated with the acid given by using the formula of depression in freezing point.
- Depression in freezing point is the decrease in the freezing point of the solvent when a non – volatile solute is added to it.
- Depression in the freezing point is given by the equation
$\Delta {{T}_{f}}=i{{K}_{f}}m$ …….(1)
where ${{K}_{f}}$is the freezing point depression constant
i is the Van’t-Hoff factor
m is the molality
Now, according to the given data we have to calculate the molality which is as follows,
\[m=\dfrac{12.2g}{122g/mol\times 100g}\times 1000g/kg=1m\]
Therefore, molality = 1m.
$\Delta {{T}_{f}}$=${{2.6}^{0}}$
${{K}_{f}}={{5.2}^{0}}kg/mol$
Substituting these values in equation (1)
\[{{2.6}^{0}}C=i\times {{5.2}^{0}}Ckg/mol\times 1mol/kg\]
\[\Rightarrow i=0.5\]
Now from this Van’t-Hoff factor i we can find the number of molecules of benzoic acid in associated state as, i =observed molecules in solution/ theoretical molecules
\[\Rightarrow i=\dfrac{1}{n}=0.5\]
\[\Rightarrow n=\dfrac{1}{0.5}=2\]
Therefore, the number of molecules of benzoic acid in the associated state will be 2.

Thus, the correct answer is option b) 2

Note: Note that depression in the freezing point is also called as cryoscopy and is also given by another formula $\Delta {{T}_{f}}=\dfrac{w}{M\times W}\times {{K}_{f}}$ where w is the weight of solute in g, W is the weight of solvent in kg and M is the molecular weight of solute. There is also another term that is the elevation in boiling point called ebullioscopy.