Question

Find the degree of association when ${\text{1}} \cdot {\text{0 gram}}$ benzoic acid, dissolved in ${\text{25 gram}}$ benzene is having depression in freezing point ${\text{0}} \cdot {\text{81k}}$. The molal depression constant for the solvent is $4 \cdot 9Kkgmol{e^{ - 1}}$.

Answer 1

Hint:Analyze the data given in the question and first find out the value of the molar mass of benzoic acid. The formula that is used for finding the molar mass of benzoic acid is $\dfrac{{1000 \times {K_f} \times {\text{Mass of benzoic acid}}}}{{\Delta {T_f} \times {\text{ Mass of solvent}}}}$. One can find out the Van’t Hoff factor of benzoic acid which is also the value of the degree of association.

Complete answer:
1) First of all, we will write the data of what has been given in the question.
Quantity of benzoic acid $ = {W_A} = 1gram$
Quantity of benzene $ = {W_B} = 25grams$
Molal depression for the reaction $ = {K_f} = 4 \cdot 9Kkgmo{l^{ - 1}}$
Depression in freezing point $ = {T_f} = 0 \cdot 81K$
The percent of the association formed $ = ?$
2) Now let us calculate the value of the molar mass of benzoic acid as below,
The molar mass of benzoic acid $ = \dfrac{{1000 \times {K_f} \times {\text{Mass of benzoic acid}}}}{{\Delta {T_f} \times {\text{ Mass of solvent}}}}$
Now by putting the values in the above formula we get,
The molar mass of benzoic acid $ = \dfrac{{1000 \times 4 \cdot 9 \times 1}}{{0 \cdot 81 \times 25}}$
Now by doing the multiplication part we get,
The molar mass of benzoic acid $ = \dfrac{{4900}}{{20 \cdot 25}}$
By calculating the above equation we get,
The molar mass of benzoic acid $ = 242{\text{ grammol}}{{\text{e}}^{ - 1}}$
3) The value we got in the above equation is abnormal i.e. experimental molar mass of benzoic acid. Now let us calculate the value of the theoretical molar mass of benzoic acid as below,
The theoretical molar mass of benzoic acid $ = {C_6}{H_6}{O_2} = 6 \times 12 + 6 \times 1 + 2 \times 16 = 122{\text{ grams}}$
4) Now let us calculate the Van’t Hoff factor which also gives us the value of the degree of association as below,
Van’t Hoff factor (i) $ = \dfrac{{{\text{Theoretical molar mass}}}}{{{\text{Experimental molar mass}}}}$
Now let us put the values in the above formula we get,
Van’t Hoff factor (i) $ = \dfrac{{122}}{{242}} = 0 \cdot 5$
Therefore, the degree of association of benzoic acid is $0 \cdot 5$.

Note:
The degree of association can be defined as the fraction of the total number of molecules which associate or combine for the formation of bigger molecules. The Van’t Hoff factor is the value of the degree of association.