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Q.

A saturated solution of a sparingly soluble salt  MCI2 has a vapour pressure of 31.78mm of Hg at 300C. While pure water exert a pressure of 31.82mm og Hg at the same temperature. Calculate the solubility product of the compound at this temp. is approximately  x × 10-5mol3L-3.  Find the value of x. (Rounding to nearest integer value)

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answer is 5.

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Detailed Solution

Let the solubility or MCI2 be small/lit
 MCI2(s)M(aq)2+2CI(aq)
Using Roult’s law
P0PsP0=nN 
31.8231.7831.82=3s100018 
On solving  s=0.0232 mol/lit
Ksp=4s3

=4.9×105=5×105

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