Question

Calculate the molality for each of the following solutions. Then, calculate the freezing-point depression ATF = iKFCm produced by each of the salts. (Assume the density of water is 1.00 g/mL and KF = 1.86°C/m.) (a) 1.8 M NaCl (Assume the density of the solution is 1.00 g/mL.) Cm = 4.9 2.0 m ATF = 49 7.5 ✓ °C (b) 24 g of KCl in 1.6 L of water cm = 49, 0.20 ATF = 49) 0.75 (c) 250 ppm MgCl2 cm = 49) 2.00 ATE = 49) 1.12 XX

Answer 1

Solution:

Part C)

Cm = 0.0026 m

ΔTf = 0.0145 °C

1 ppm = 1 mg / L

Thus,

250 ppm = 250 mg / L = 0.250 g / L

Since,

Molality (Cm) = Mass /Molar mass x Mass of water in kg

(Since, density of water = 1 g/mL

Hence, volume of solution = mass of solution

Mass of solution = mass of water + mass of MgCl2

Therefore,

Mass of water (solvent) = 1000 g - 0.250 g = 999.75 g )

Molality = 0.250 g / 95.21 g mol-1 x 0.999 g

C m = 0.0026 m

Since,

ΔTf = i Kf Cm

i for MgCl2 = 3 (due to one Mg2+ and two Cl- ions)

.ΔTf = 3 x 1.86 °C/m x 0.0026 m

= 0.0145 °C