`K_(sp)` of `AgCl` is

What is the K_(sp) of AgCl if it is given that solubility of AgCl at 20^(@)C is 1.435xx10^(-5)g//L ?

K_(sp) of AgCl in water at 25^(@)C is 1.8xx10^(-10) . If 10^(-5) mole of Ag^(+) ions are added to this solution. K_(sp) will be:

K_(sp) of AgCl in water at 25^(@)C is 1.8xx10^(-10) . If 10^(-5) mole of Ag^(+) ions are added to this solution. K_(sp) will be:

Calculate the ratio of solubility of AgCl " at " 25 ^(@) C in 3.0 M NH_3 (K_(sp) of AgCl in NH_3 " is " 3.1 xx 10^(-3) ) and " in " H_2O (K_(sp) of AgCl in H_2O " is" 1.8 xx 10^(-10)) . If answer is 1.15 xx 10 ^(x) then x=________?

The solubility of AgCl in 0.1 M NaCl is ( K_(sp) of AgCl = 1.2 xx 10^(-10) )

The solubility of AgCl in 0.1M NaCI is (K_(sp) " of AgCl" = 1.2 xx 10^(-10))

The solubility of AgCl in 0.1M NaCI is (K_(sp) " of AgCl" = 1.2 xx 10^(-10))

If the solubility of AgCl in 0.1 M NaCl is ( K_(sp) of AgCl = 1.2 xx 10^(-10) )