SolutionHard
Question
When 100 ml of 0.2M BaCl2 (aq) is added to 100ml of 0.2M K2SO4 solution, of K2SO4 solution is lowered. Assume 100% dissociation of soluble salts and 100% precipitation of insoluble ones.
Options
A.Freezing point
B.Boiling point
C.Vapour pressure
D.cc
Solution
K2SO4 + BaCl2 → BaSO4 + 2KCl
0.02 0.02 0 0
0 0 0.02 0.04
ᐃTb = m Kbi
(ᐃTb)i = (0.02 × 2 + 0.02) × Kb = 0.06 Kb
(ᐃTb)final = 0.04 × 2 × Kb = 0.08 Kb
ᐃTb ↑ So B.P. ↑
(ᐃTf)i = - 0.06 Kf ; (ᐃTf)f = 0.08 Kf
ᐃTf ↑ So F.P. ↓
= Xsolute ; Xsolute ↑ So dP ↓ VP ↓
πinitial = M R T i = 0.06 RT
πfinal = 0.08 RT
So, π↑
0.02 0.02 0 0
0 0 0.02 0.04
ᐃTb = m Kbi
(ᐃTb)i = (0.02 × 2 + 0.02) × Kb = 0.06 Kb
(ᐃTb)final = 0.04 × 2 × Kb = 0.08 Kb
ᐃTb ↑ So B.P. ↑
(ᐃTf)i = - 0.06 Kf ; (ᐃTf)f = 0.08 Kf
ᐃTf ↑ So F.P. ↓
= Xsolute ; Xsolute ↑ So dP ↓ VP ↓πinitial = M R T i = 0.06 RT
πfinal = 0.08 RT
So, π↑
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