JEE Main 2019 — Ionic Equilibrium Question with Solution
From: JEE Main 2019 (Online) 10th January Morning Slot
Question
A mixture of 100 m mol of Ca(OH)2 and 2 g of sodium sulphate was dissolved in water and the volume was made up to 100 mL. The mass of calcium sulphate formed and the concentration of OH– in resulting solution, respectively, are : (Molar mass of Ca (OH)2, Na2SO4 and CaSO4 are 74, 143 and 136 g mol–1
, respectively; Ksp of Ca(OH)2 is 5.5 × 10–6
)
Choose an option
Show full solutionCorrect option: C
Correct answer
C1.9g, 0.28 mol L1
Step-by-step explanation
Ca(OH)2 + Na2SO4 CaSO4 + 2NaOH
100 m mol 14 m mol
14 m mol 28 m mol
wCasO4 = 14 103 136 = 1.9 gm
[OH] = = 0.28 M
100 m mol 14 m mol
14 m mol 28 m mol
wCasO4 = 14 103 136 = 1.9 gm
[OH] = = 0.28 M
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