Question

if 95.21 g of MgCl2, a strong electrolyte is added to 650 g of water,how much would the freezing point of water be lowered? What would be freezing point of this solution?The Kf for water is 1.86 degrees C/m.

Answer 1

Answer: Freezing point of this solution is $-8.58^0C$

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

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

i= vant hoff factor = 3 (for $MgCl_2$ as it dissociates to give three ions.

$K_f$ = freezing point constant = $1.86^0C/m$

m= molality

Weight of solvent (water)= 650g = 0.65 kg

Molar mass of solute = 95.21 g

Molar mass of solute= 95.21 g/mol

$0-T_f=3\times 1.86\times \frac{\95.21g}{95.21g/mol}\times 0.65kg$

$T_f=-8.58^0C$

Thus freezing point of this solution is $-8.58^0C$