Question

Preparing a solution of known freezing temperature:

Assigned temperature= -2 degrees celsius

Mass: 10 mL

Calulate the appropiate mass of NaCl?

Answer 1

mass of water = volume of water . because density is 1 g / mL

mass of solvent water = 10 g

i value for NaCl = 2 (because two ions Na+ and Cl- )

dT = freezing point depression = To - Tf

dT = 0 -(-2) = 2 oC

Kf of water = 1.86 oC / m

dT = i x Kf x molality

2 = 2 x 1.86 x molality

molality   = 0.538 m

molality = moles / mass of solvent (kg)

0.538 = moles / 0.01

moles = 5.38 x 10^-3

moles of NaCl = mass of NaCl / molar mass of NaCl

5.38 x 10^-3 = mass / 58.5

mass = 0.32 g

mass of NaCl = 0.32 g