Question

A0.5% aqueous solution of kcl is found to freeze at -0.24c. Calculate the van't hoff factor of solute at this concentration.(k1 for water is 1.86for1000g)

Answer 1

Van't hoff factor of solute at this concentration is 1.8

Explanation:

Depression in freezing point is given by:

$\Delta T_f=K_f\times m$

$\Delta T_f=T_f^0-T_f=(0-(-0.24)^0C=0.24^0C$ = Depression in freezing point

$K_f$ = freezing point constant = $1.86K/kgmol$

m= molality

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

0.5 g of KCl is dissolved in 100 g of solution.

Thus mass of solvent = mass of solution - mass of solute = 100-0.5 = 99.5 g = 0.0995 kg    (1kg=1000g)

$0.24^0C=i\times 1.86\times \frac{0.5g}{74.5g/mol\times 0.0955kg}$

$0.24=i\times 0.13$

$i=1.8$

Thus van't hoff factor of solute at this concentration is 1.8

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