Question

A solution containing 0.5 g KCl dissolved in 100g of water freezes at-0.23°C. calculate the degree of dissociation of salt kf for water=1.86kper m

Answer 1

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Answer 2

Answer: 84%

Explanation:-

Weight of solvent 100 g = 0.1 kg (1 kg=1000 g)

Molar mass of KCl = 74.5 g/mol

Mass of KCl added = 0.5 g

$\Delta T_f=i\times K_f\times \frac{\text{mass of KCl}}{\text{molar mass of KCl}\times \text{weight of solvent in kg}}$

i = Van'T Hoff factor

$\Delta T_f=T^{o}_f-T_f=0-(-0.23)^0C$

$\Delta T_f=0.23K$

$0.23=i\times 1.86 K kg/mol\frac{0.5 g}{74.5g/mol\times 0.1kg}$

$i=1.84$

$\alpha =\frac{i-1}{n-1}$

$KCl\rightarrow K^++Cl^-$

$\alpha =\frac{1.84-1}{2-1}=0.84$

Degree of dissociation is 84%.