Determine the freezing point depression of a solution that contains 30.7 g glycerin (c3h8o3, molar mass = 92.09 g/mol) in 376 ml of water. some possibly useful constants for water are kf = 1.86°c/m and kb = 0.512°c/m.

Answer : The freezing depression of a solution is

Explanation :  Given,

Molal-freezing-point-depression constant for water =

Mass of glycerin (solute) = 30.7 g

Volume of water (solvent) = 376 mL

Molar mass of glycerin = 92.09 g/mole

First we have to calculate the mass of water.

As we know that, the density of water is 1.0 g/mL.

So, mass of water = Density of water × Volume of water

Mass of water = 1.0 g/mL × 376 mL

Mass of water = 376 g = 0.376 kg   (1 g = 0.001 kg)

Now we have to calculate the freezing point depression of a solution.

Formula used :  

where,

= change in freezing point

= freezing point of solution = ?

= freezing point of water =

i = Van't Hoff factor = 1  (for glycerin non-electrolyte)

= freezing point constant for water =

m = molality

Now put all the given values in this formula, we get

Therefore, the freezing depression of a solution is