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.

Freezing point of this solution is

Explanation:

Depression in freezing point is given by:

= Depression in freezing point

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

= freezing point constant =

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

Thus freezing point of this solution is

- lead + phosphoric acid = hydrogen + lead(ii) phosphate

the answer is 0.42 is that is an option. this question provides no info