Question

Molal Boiling-Point-Elevation and Freezing-Point-Depression
Solvent	Normal Boiling Point (∘C)	Kb (∘C/m)	Normal Freezing Point (∘C)	Kf (∘C/m)
Water, H2O	100.0	0.51	0.0	1.86
Benzene, C6H6	80.1	2.53	5.5	5.12
Ethanol, C2H5OH	78.4	1.22	-114.6	1.99
Carbon tetrachloride, CCl4	76.8	5.02	-22.3	29.8
Chloroform, CHCl3	61.2	3.63	-63.5	4.68

Part E

freezing point of 2.02 g KBr and 4.84 g glucose (C6H12O6) in 187 g of water

Part F

boiling point of 2.02 g KBr and 4.84 g glucose (C6H12O6) in 187 g of water

Part G

freezing point of 1.40 g NaCl in 0.255 kg of water

Part H

boiling point of 1.40 g NaCl in 0.255 kg of water

Answer 1

Part E : Freezing point = -0.605 ^oC

Part F : boiling point = 100.166 ^oC

Part G : Freezing point = -0.350 ^oC

Part H : boiling point = 100.096 ^oC

Explanation

Part G

mass NaCl = 1.40 g

moles NaCl = (mass NaCl) / (molar mass NaCl)

moles NaCl = (1.40 g) / (58.44 g/mol)

moles NaCl = 0.024 mol

molality NaCl = (moles NaCl) / (mass of water in kg)

molality NaCl = (0.024 mol) / (0.255 kg)

molality NaCl = 0.094 m

decrease in freezing point = (i) * (K_f) * (molality NaCl)

decrease in freezing point = (2) * (1.86 ^oC/m) * (0.094 m)

decrease in freezing point = 0.350 ^oC

freezing point of solution = (freezing point of pure solvent) - (decrease in freezing point)

freezing point of solution = (0.0 ^oC) - (0.350 ^oC)

freezing point of solution = -0.350 ^oC