1.5 g of a monobasic acid when dissolved in 150 g of water lowers the freezing point by 0.165° 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 (Kf for water =1.86° C mol -1 ).

Question

1.5 g of a monobasic acid when dissolved in 150 g of water lowers the freezing point by 0.165° 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 (Kf for water =1.86° C mol -1 ). (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

((0.5/ M ))=((37.5/10)) × 10-3 M =(5/37.5) × 1000 M =133.33 gm Δ T f = iKf textrm m 0.165=( i × 1.86 × 1.5 / 133.3/150 / 1000) i =1.182 α=( i -1/ n -1) n =2 α=18.28 %

Q. 1.5g of a monobasic acid when dissolved in 150g of water lowers the freezing point by 0.165∘C.0.5g of the same acid when titrated, after dissolution in water, requires 37.5ml of N/10 alkali. Calculate the degree of dissociation of the acid (Kf​ for water =1.86∘Cmol−1 ).

(M0.5​)=(1037.5​)×10−3 M=37.55​×1000 M=133.33gm ΔTf​=iKf​m 0.165=150/1000i×1.86×1.5/133.3​ i=1.182 α=n−1i−1​{n=2} α=18.28%

$(\frac{0.5}{M}) = (\frac{37.5}{10}) \times 1 0^{- 3}$
$M = \frac{5}{37.5} \times 1000$
$M = 133.33 g m$
$Δ T_{f} = i K_{f} m$
$0.165 = \frac{i \times 1.86 \times 1.5/133.3}{150/1000}$
$i = 1.182$
$α = \frac{i - 1}{n - 1} {n = 2}$
$α = 18.28%$