Answer:
0.640 Torr; 1.28 Torr
Step-by-step explanation:
According to Raoult's Law,
(1) p₁ = χ₁p₁°,
where p₁ is the solvent
χ₁ + χ₂ = 1 Subtract χ₁ from each side
χ₂ = 1 – k₁ Substitute in (1)
(2) p₁ = (1 – χ₂)p₁° Remove parentheses
p₁ = p₁°- χ₂p₁° Subtract p₁° from each side
p₁ - p₁° = -χ₂p₁° Multiply each side by -1
Δp = χ₂p₁°
Δp is the vapour pressure lowering.
If the solute is an electrolyte, we must insert the van't Hoff i factor.
ΔP = iχ₂P₁°
===============
(1) Calculate χ₂ for each solution
Water:
1000 g = 1000 × 1/18.02
= 55.49 mol
χ₂ = 2.10/(2.10 + 55.49)
= 2.10/57.59
= 0.036 46
===============
(2) Calculate the vapour pressure lowering
(a) Sucrose
i = 1
ΔP = 1 × 0.036 46 × 17.54
ΔP = 0.640 Torr
=====
(b) NaCl
i = 2
ΔP = 2 × 0.640
ΔP = 1.28 torr
