AP EAMCET202022 Sep 2020Morning ShiftChemistryIonic EquilibriumActual
K _ sp for CaSO ₄ is 9 10⁻⁶ . The minimum volume of water needed to dissolve 1 ~g of CaSO ₄ at 298 ~K temperature is ......
Options
- A3.50 ~L
- B4.25 ~L
- C1.75 ~L
- D2.45 ~L
Correct answer
D. 2.45 ~L
Step-by-step solution
Let S be the solubility of CaSO ₄ . aligned & CaSO ₄ Ca ²⁺+ SO ₄²⁻ & [ Ca ²⁺ ]= [ SO ₄²⁻ ]=S & K_ sp [ Ca ²⁺ ] [ SO ₄²⁻ ]=S S=S^2=9 10⁻⁶ & S=0.003 M aligned The molar mass of CaSO ₄ is 40+32+64=136 ~g The solubility in g / L is 0.003 136=0.408 ~g / L means, 0.408 ~g dissolves in 1 ~L . 1 ~g will dissolve in =1 / 0.408=2.45 ~L Hence, the correct option is (4).