question_answer

0.400 g of an acid HA (mol. mass = 80) was dissolved in 100 g of water. The solution showed a depression of freezing point of 0.12 K. What will be the dissociation constant (in multiple of \[{{10}^{-\,3}}\]) of the acid at about \[0{}^\circ C\]? Given \[{{K}_{f}}(water)=1.86\text{ }K\text{ }Kg\text{ }mo{{l}^{-\,1}}\] (Assume molarity of solution \[\approx \] molality)

Answer:

5.92