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6-2.Equilibrium-II (Ionic Equilibrium)
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જો $K_{sp}\, 0.001435$ ગ્રામ $AgCl\, 100\, ml$ પાણીમાં હાજર હોય તો $k_{sp}$ ગણો.

A

$10^{-6}$

B

$10^{-8}$

C

$10^{-10}$

D

$10^{-5}$

Solution

$AgCl$ માં $100\, ml $  $= 0.001435\, g$

$1000 \,ml = 0.001435 \times 10 = 0.01435\, g $

ગ્રામ મોલ $  = \,\frac{{{\text{0}}{\text{.001435}} \times {\text{10}}}}{{143.5}} = {10^{ – 4}}\,gram\,\,mole/litre$

$K_{sp} = s^2 = (10^{-4}) = 10^{-8}$ ગ્રામ મોલ/લીટર

Standard 11
Chemistry

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