Answer:- 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.