What mass of salt (nacl) should you add to 1.94 l of water in an ice cream maker to make a solution that freezes at -14.4 ∘c ? assume complete dissociation of the nacl and density of 1.00 g/ml for water.

- mass of NaCl used is 439g.

Solution:- Depression in freezing point is a colligative property and the formula used to solve this type of problems is,

delta Tf = i x m x kf

where, delta Tf is depression in freezing point

i is Van't hoff factor. For NaCl, the value of i is 2 as it gives two ions.

m is molality and kf is the molal freezing point constant. For water, kf is 1.86 degree C/m

given, Volume of water = 1.94 L = 1.94 kg

Freezing point of solution = -14.4 degree C

pure water(solvent) freezes at 0 degree C. So, delta Tf = 0 - (-14.4) = 14.4 degree C

Let's plug in the values in the formula..

14.4 = 2 x m x 1.86

m = 14.4/(2 x 1.86)

m = 3.87

molality, m = moles of solute/kg of solvent

moles of solute = m x kg of solvent = 3.87 x 1.94 = 7.51

mass of solute, NaCl used = 7.51 mol x (58.44 g/mol) = 438.88 g that could be round to 439g.

hi my dude or dudette

frequency, z, and the collision density, z, in carbon dioxide (co2), r 180 pm at 30 °c and 120 kpa