The solubility product constant of calcium sulfate, caso4, is 7.10×10−5. its molar mass is 136.1 g/mol. how many grams of calcium sulfate can dissolve in 41.5 l of pure water

the mass of calcium sulfate that can be dissolved  = 47.64 grams

Explanation:

Ksp of CaSO₄ = 7.10X10⁻⁵

The expression for Ksp will be:

Ksp = [Ca⁺²][SO₄⁻²]

Let solubility of CaSO₄ = s

Therefore

[Ca⁺²] = [SO₄⁻²]=s

Ksp = s² = 7.10X10⁻⁵

s = 8.43 X10⁻³ mol/L

this is the solubility thus this is the moles of calcium sulfate that can be dissolved per litre of pure water.

So the moles that can be dissolved in 41.5 L of water

                       =  8.43 X10⁻³X41.5 = 0.35 moles

the mass of calcium sulfate that can be dissolved = moles X molar mass

the mass of calcium sulfate that can be dissolved = 0.35X136.1 = 47.64 grams

idk what the question is about

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole