72) what is the freezing point (°c) of a solution prepared by dissolving 11.3 g of ca(no3)2 (formula weight = 164 g/mol) in 115 g of water? the molal freezing point depression constant for water is 1.86 °c/m. g

The freezing point of solution is -3.34°C

Explanation:

Vant hoff factor for ionic solute is the number of ions that are present in a solution. The equation for the ionization of calcium nitrate follows:

The total number of ions present in the solution are 3.

Where,

= Given mass of solute = 11.3 g

= Molar mass of solute = 164  g/mol

= Mass of solvent (water) = 115 g

Putting values in above equation, we get:

where,

i = Vant hoff factor = 3

= molal freezing point depression constant = 1.86°C/m.g

m = molality of solution = 0.599 m

Putting values in above equation, we get:

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

= 3.34 °C

Freezing point of water = 0°C

Freezing point of solution = ?

Putting values in above equation, we get:

Hence, the freezing point of solution is -3.34°C

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y