Question

1.5 g of a monobasic acid, when dissolved in 150g of water, lowers the freezing point by ${0.165}^{\circ }$C. 0.5 g of the same acid when titrated, after dissolution in water, requires 37.5 ml of N/10 alkali. Calculate the degree of dissociation of the acid ?($K_f$for water =${1.86}^{\circ }$C mol${}^{-1}).$

• A. $11.34$%
• B. $34.64$%
• C. $65.36$%
• D. $89.64$%

Answer 1

The correct option is A $11.34$%

$\frac{0.5}{M}=3.75\times {10}^{-3}$

$M=133.33$

$0.165=(1+\alpha )\times 11.86\times 1.5\times \frac{1000}{133.33}\times 150$

$\alpha =0.1827=18.27$%