SolutionHard
Question
25 mL of an aqueous solution of KCl was found to requires 20 mL of 1M AgNO3 solution when titrated using a K2CrO4 as indicator. Depression in freezing point of KCl solution with 100% ionization will be
[Kf = 2.0o mol-1 kg and molarity = molality]
[Kf = 2.0o mol-1 kg and molarity = molality]
Options
A.20/45
B.80/45
C.40/45
D.160/45
Solution
KCl + AgNO3 → AgCl + KNO3
0.8 × 25 1 × 20 0 0
(mili mole) (mili mole)
0 0 20 20
Mole of KNO3 in solutions =
ᐃTf = i × m × Kf
ᐃTf = 2 ×
× 2 = 
0.8 × 25 1 × 20 0 0
(mili mole) (mili mole)
0 0 20 20
Mole of KNO3 in solutions =
ᐃTf = i × m × Kf
ᐃTf = 2 ×
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