The solubility of AgCl of water at 25- C is 1-79 - 10-3 g - lit- Calculate K sp AgCl at 25- C- A- 3-45 - 10-10B- 4-5 - 10-10C- 1-55 - 10-10D- 9-7 - 10-10

The correct option is C 1.79×10^ 3g/lit =1.79×10^ 3/143.5 mole/lit =1.247× 10^ 5 mole/lit ( K_sp )AgCl= [ Ag^+ ] [ Cl ]=S^2 = ( 1.247× 10^ 5 )^2=1.55×10^ 10 ...

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Standard XII

Chemistry

Percentage Composition

The solubilit...

Question

The solubility of AgCl of water at $25^{\circ}$ C is $1.79 \times 10^{- 3}$ g/lit. Calculate $(K_{s p})$ AgCl at $25^{\circ} C .$

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Solution

The correct option is C
$1.79 \times 10^{- 3} g / l i t$
$= \frac{1.79 \times 10^{- 3}}{143.5}$ mole/lit
$= 1.247 \times 10^{- 5}$ mole/lit
$(K_{s p}) A g C l = [A g^{+}] [C l] = S^{2}$
$= {(1.247 \times 10^{- 5})}^{2} = 1.55 \times 10^{- 10}$

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The solubility of AgCl of water at 25∘ C is 1.79×10−3 g/lit. Calculate (Ksp) AgCl at 25∘C.

The correct option is C 1.79×10−3g/lit =1.79×10−3143.5 mole/lit =1.247×10−5 mole/lit (Ksp)AgCl=[Ag+][Cl]=S2 =(1.247×10−5)2=1.55×10−10

The solubility of AgCl of water at $25^{\circ}$ C is $1.79 \times 10^{- 3}$ g/lit. Calculate $(K_{s p})$ AgCl at $25^{\circ} C .$

The correct option is C
$1.79 \times 10^{- 3} g / l i t$
$= \frac{1.79 \times 10^{- 3}}{143.5}$ mole/lit
$= 1.247 \times 10^{- 5}$ mole/lit
$(K_{s p}) A g C l = [A g^{+}] [C l] = S^{2}$
$= {(1.247 \times 10^{- 5})}^{2} = 1.55 \times 10^{- 10}$