Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • What is the minimum volume of water required to dissolve 1g of calcium sulphate at 298 K? (For calcium sulphate, K_sp is 9.1 × 10^–6)

7.72     What is the minimum volume of water required to dissolve 1g of calcium sulphate at 298 K? (For calcium sulphate, Ksp is 9.1 × 10-6)

Answers (1)

best_answer

We have,
The solubility product of calcium sulphate is 9.1\times 10^{-6}.
given mass of calcium sulphate = 1g

Ionization of calcium sulphate;

CaSO_4\rightleftharpoons Ca^{2+}+SO_{4}^{2-}
Therefore, 
K_s_p = [Ca^{2+}][SO_{4}^{2-}]

Let the solubility of calcium sulphate be s.
Then, 
\\K_{sp} = s^2\\ 9.1\times 10^{-6}=s^2\\ s = \sqrt{9.1\times 10^{-6}}
     =3.02\times 10^{-3} mol/L

Thus,
mass/ (mol. wt)\times volume =3.02\times 10^{-3} Molarity
mass = 3.02\times 136\times 10^{-3} = 0.41 g

So, that to dissolve 1 g of calcium sulphate we need = 1/0.41 L = 2.44 L of water.

Posted by

manish

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Class 12 board exams 2025 to have 50% competency-based questions; no change in 10th
anam-khan

CBSE Class 12 board exams 2024 over; expected result date, trends of last 10 years

Latest Question