according to this equation:
AgCl(s) ↔ Ag+(aq) + Cl-(aq)
so K1 = [Ag+][Cl-]
when [Ag+] = [Cl-] we can assume both = X
and when we have X the solubility = 1.33 x 10^-5 mol / L
by substitution:
∴ K1 = X^2
= (1.33 x 10^-5)^2
= 1.77 x 10^-10
by using vant's Hoff equation:
ln(K2/K1) = (ΔH/R)*(1/T2-1/T1)
when ΔH = 65700 J / mol
R = 8.314
T1 = 25+273 = 298 K
T2 = 47.7 +273 =320.7
by substitution:
∴㏑(K2/1.77 x 10^-10) = (65700/8.314) * ( 1/320.7 - 1/ 298)
by solving for K2
∴K2 = 2.7 x 10^-11
and when K2 = X^2
∴ the solubility X = √(2.7 x 10^-11)
= 5.2 x 10^-6 mol/L