Chemistry - Liquid Solution Question with Solution | TestHub

From titration:

Moles of acid = $\frac{N\times V}{1000}=\frac{0.1\times 37.5}{1000}=0.00375$

For 0.5 g acid:

$M=\frac{0.5}{0.00375}=133.3$

For 1.5 g acid:

moles = $\frac{1.5}{133.3}=0.01125$

Molality:

$m=\frac{0.01125}{0.150}=0.075$

$\Delta T_f=i{K}_{f}m$

$0.165=i(1.86)(0.075)$

$i=1.18$

For monobasic acid:

$i=1+\alpha$

$\alpha =1.18-1=0.18$

Therefore, the degree of dissociation = 0.18 or 18%.