Question

A solution containing 3.10g of BaCl2 in 250g of water, boils at 100.083C.Calculate molality and van't Hoff factor of BaCl2 in this solution (Kb=0.52 K kg mol-,molar mass of Ba(No3)2=208.3 g/mol)

Answer 1

molality is 0.015 and vant haff factor is 2.8

Answer 2

Answer:

The value of molality of the solution is 0.0595 mol/kg.

The van't Hoff factor of $BaCl_2$ in this solution is 2.68.

Explanation:

Mass of $BaCL_2$ = 3.10 g

Mass of water = 250 g = 0.250 kg

molality of the solution ,m=$\frac{3.10 g}{208.3 g/mol\times 0.250 kg}=0.0595 mol/kg$

Boiling point of the solution = $T_b=100.083^oC=373.083K$

Boiling poiunt of water = $T=100^oC=373 K$

$\Delta T_b=T_b-t=373.083K-373K=0.083 K$

$\Delta T_b=ik_f\times m$

$0.083 K=i\times 0.52 K kg/mol\times 0.0595 mol/kg$

i = 2.68

The value of molality of the solution is 0.0595 mol/kg.

The van't Hoff factor of $BaCl_2$ in this solution is 2.68.