Solubility of 16 × {10^{ - 4}},m/s at {20,^o}C is 1.435 × {10^{ - 3}},gm,per,litre . solubility pr

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6-2.Equilibrium-II (Ionic Equilibrium)

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Solubility of $16 \times {10^{ - 4}}\,m/s$ at ${20\,^o}C$ is $1.435 \times {10^{ - 3}}\,gm\,per\,litre$. The solubility product of $AgCl$ is

A

$C{O_2}$

B

$1 \times {10^{ - 10}}$

C

$1.435 \times {10^{ - 5}}$

D

$108 \times {10^{ - 3}}$

(AIPMT-2002)

Solution

(b) $S = 1.435 \times {10^{ – 3}}\,g/l$,$ = \frac{{1.435 \times {{10}^{ – 3}}}}{{143.5}} = {10^{ – 5}}$ $M$

${K_{sp}} = S \times S = {10^{ – 10}}$

Standard 11

Chemistry

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