A 0.01\; m aqueous solution of AlCl_{3} freezes at -0.068^{\circ}C. Calculate the percentage of dissociation. [Given : K_{f} for Water=1.86\; K\; kg\; mol^{-1}]

 

Answers (1)
S safeer

ΔTf = i(Kf)m

0.068 = i x 1.86 x 0.01

i = 3.65 or 3.656

α = i-1/n-1

α = 0.883 or 0.885

Hence percentage of dissociation is 88.3% or 88.5%

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