JEE Mains · Chemistry · STD 11 - 6.2. Equilibrium - II (icon Equilibrium)
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 upto \(100\, mL\). The mass of calcium sulphate formed and the concentration of \(OH^-\) in resulting solution, respectively are (Molar mass of \(Ca(OH)_2, Na_2SO_4\) and \(CaSO_4\) are \(74, 143\) and \(136\, g\, mol^{-1}\) respectively; \(K_{sp}\) of \(Ca(OH)_2\) is \(5.5 \times 10^{-6}\))
- A \(1.9\, g, 0.28\, mol\, L^{-1}\)
- B \(13.6\, g, 0.28\, mol\, L^{-1}\)
- C \(1.9\, g, 0.14\, mol\, L^{-1}\)
- D \(13.6\, g, 0.14\, mol\, L^{-1}\)
Answer & Solution
Correct Answer
(A) \(1.9\, g, 0.28\, mol\, L^{-1}\)
Step-by-step Solution
Detailed explanation
\(\mathop {\mathop {Ca{{(OH)}_2}}\limits_{100\,m\,mol} }\limits_ - + \mathop {\mathop {N{a_2}S{O_4}}\limits_{14\,m\,mol} }\limits_ - \to \mathop {\mathop {CaS{O_4}}\limits_ - }\limits_{14\,m\,mol} + \mathop {\mathop {2NaOH}\limits_ - }\limits_{28\,m\,mol} \)…
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