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Depression of freezing point of. 01 molal aqueous Acetic Acid solution is. 02 04 6 degree Celsius .1 molal urea freezes at minus point - 1.86 degree Celsius. assuming molality equal to molarity pH of acetic acid is

Depression of freezing point of. 01 molal aqueous Acetic Acid solution is. 02 04 6 degree Celsius .1 molal urea freezes at minus point - 1.86 degree Celsius. assuming molality equal to molarity pH of acetic acid is

Grade:12

1 Answers

Arun
25750 Points
6 years ago
Depression of freezing pointTf = Kf × i× m1 molal urea solution freezes at 1.86 oC1.86 = Kf × 1Kf =1.86Using the value of Kf, find the i valueTf = Kf × m0.0204 = 1.86 ×i× 0.01i= 1.1CH3COOH CH3COO +H+C(1α)                   Cα               Cαi= 1 +αα = 0.1[H+] = Cα = 0.01 × 0.1 = 0.001pH =  log [H+]    =  log 0.001   = log (1 × 103)    =3

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