Chemistry(A) 0.1M AgNO3 (B) 0.1M CaCl2 (C ) 0.15M NaCl (D) H2OCorrect decreasing trend of solubility of AgCl in (A), (B), (C) and (D) will be

$(A) 0 .1 M {AgNO}_{3} (B) 0 .1 M {CaCl}_{2} (C) 0 .15 M NaCl (D) H_{2} O$
Correct decreasing trend of solubility of AgCl in (A), (B), (C) and (D) will be

A
B > A > C > D
B
D > A = B > C
C
C > A = B > D
D
D > A > C > B

Solution:

$Solubility \propto \frac{1}{Common ion concentration} lons present in AgCl are {Ag}^{+} and {Cl}^{-} (A) [{Ag}^{+}] = 0 .1 M \times 1 (B) [{Cl}^{-}] = 0 .1 M \times 2 (C) [{Cl}^{-}] = 0 .15 M \times 1 (D) No common ion ∴ Solubility of AgCl in D > A > C > B$

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