The solubility of agcl(s) in water at 25 degrees celsius is 1.33 x 10^-5 mol/l and its delta h of solution is 65.7 kj/mol. what is its solubility at 45 degrees celsius?
express your answer using two significant figures.
s_{\rm agcl} =
\rm m
find molarity of agcl(s) at 45 degrees celsius?

S AgCl(45°C) = 3.09 E-5 mol/L

Explanation:

          S             S          S

∴ S = 1.33 E-5 mol/L

⇒Ksp1 (25°C) = S² = 1.8 E-10

Vant Hoff:

∴ T1 = 25°C = 298.15K

∴T2 = 45°C = 318.15 K

∴ R = 8.314 E-3 KJ/mol.K

⇒ K1/K2 = e ( -ΔH/R (1/T1 - 1/T2))

⇒ K1/K2 = e ( - (65.7/8.314E-3) ( 1/298.15 - 1/318.15))

⇒ K1/K2 = e ( - 1.666 )

⇒ K1/K2 = 0.1889

⇒ K2 = K1/0.1889 = 1.8 E-10 / 0.1889 = 9.525 E-10 (45°C)

⇒ Ksp(45°C) = S² = 9.525 E-10

⇒ S = √9.525 E-10 = 3.0863 E-5 mol/L

Molarity (M):

M ≡ mol/L

M AgCl = S = 3.09 E-5 M