The solubility of agcl(s) in water at 25°c is 1.33 x 10-5 mol/l and its δh° of solution is 65.7 kj/mol. what is its solubility at 95°c?

The solubility of agcl(s) in water at 95°C is : 5.4 * 10⁻⁴  mol/L

Given data :

solubility of agcl(s) at 298 k ( 25°c ) = 1.33 * 10⁻⁵ mol/l

ΔH∘ of solution =  65.7 kJ/mol

applying van't hoff equation

ln  -  ΔH∘ / R * ( 1 / T₂ - 1 / T₁ )  --- ( 1 )

where : T₂ = 273 + 95 = 368 k , T₁ = 298 k

Substitute values into equation ( 1 )

ln ( Ksp / 1.7689 * 10⁻¹⁰ ) =  - 65.7 * 10³ / 8.314 * ( 1 / 368 - 1 / 298 )

therefore :

( Ksp / 1.7689 * 10⁻¹⁰ ) = -7.90 * 10³ * -0.00064

                                    = e^5.056

therefore:  Ksp = 1.7689 * 10⁻¹⁰ * 156.96

                         = 2.776 * 10⁻⁸

Given that:   Ksp = S² ( solubility )

solubility ( s ) at 95° C  = 5.4 * 10⁻⁴  mol/L

                                     = 5.4 * 10⁻⁴  mol/L

Hence we can conclude that The solubility of agcl(s) in water at 95°C is : 5.4 * 10⁻⁴  mol/L

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i think   the answer is c hope i'm right

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2