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Find K sp of CaCl 2 , if Λ m c =x Scm 2 mol -1 and K (kappa )=0.001 S cm -1 at 298 K

Find Ksp of CaCl2 , if Λmc=x Scm2 mol-1 and K (kappa )=0.001 S cm -1 at 298 K

Grade:12

1 Answers

Bhavya
askIITians Faculty 1281 Points
7 years ago
We know that
\Lambda= (K * 1000)/ solubility
Applying the values we get,
solubility = (0.001 * 1000)/x = 1/x
Now,
CaCl2 ---------> Ca+2 + 2Cl-
s 2s
Ksp = s*(2s)2 =4s3
= 4/x3

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