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A mixture of 100 m mol of Ca(OH)2 and 2g 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)
  • a)
    1.9 g, 0.14 mol L–1
  • b)
    13.6 g, 0.14 mol L–1
  • c)
    1.9 g, 0.28 mol L–1
  • d)
    13.6 g, 0.28 mol L–1
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
A mixture of 100 m mol of Ca(OH)2 and 2g of sodium sulphate was dissol...
Ca(OH)2  + Na2SO4 → CaSO4 + 2NaOH 100 m mol 14 m mol         —    — —     — 14 m mol 28 m mol
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Most Upvoted Answer
A mixture of 100 m mol of Ca(OH)2 and 2g of sodium sulphate was dissol...
To determine the mass of calcium sulfate formed, we need to calculate the number of moles of calcium sulfate produced.

First, we need to determine the limiting reactant. The reaction between calcium hydroxide (Ca(OH)2) and sodium sulfate (Na2SO4) can be represented as follows:

Ca(OH)2 + Na2SO4 -> CaSO4 + 2NaOH

The molar mass of Ca(OH)2 is 74.09 g/mol, and the molar mass of Na2SO4 is 142.04 g/mol.

1. Calculate the number of moles of calcium hydroxide (Ca(OH)2):
100 mmol = 100/1000 = 0.1 mol

2. Calculate the number of moles of sodium sulfate (Na2SO4):
mass = 2 g
molar mass = 142.04 g/mol
moles = mass/molar mass = 2/142.04 = 0.0141 mol

3. Determine the limiting reactant:
Since the mole ratio between Ca(OH)2 and Na2SO4 is 1:1, and we have less moles of Na2SO4 (0.0141 mol) compared to Ca(OH)2 (0.1 mol), Na2SO4 is the limiting reactant.

4. Calculate the number of moles of calcium sulfate produced:
The mole ratio between Na2SO4 and CaSO4 is 1:1, so the number of moles of CaSO4 produced is also 0.0141 mol.

5. Calculate the mass of calcium sulfate:
The molar mass of CaSO4 is 136.14 g/mol.
mass = moles * molar mass = 0.0141 * 136.14 = 1.92 g

Therefore, the mass of calcium sulfate formed is 1.92 g.

To determine the concentration of OH- ions in the solution, we need to calculate the number of moles of OH- ions and then divide by the volume of the solution.

1. Calculate the number of moles of OH- ions:
From the balanced chemical equation, we can see that 1 mol of Na2SO4 produces 2 moles of NaOH. Therefore, the number of moles of NaOH produced is 2 * 0.0141 mol = 0.0282 mol.

2. Calculate the concentration of OH- ions:
The volume of the solution is 100 mL, which is equal to 100/1000 = 0.1 L.
Concentration = moles/volume = 0.0282 mol / 0.1 L = 0.282 M

Therefore, the concentration of OH- ions in the solution is 0.282 M.
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A mixture of 100 m mol of Ca(OH)2 and 2g 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; Kspof Ca(OH)2 is 5.5 × 10–6)a)1.9 g, 0.14 mol L–1b)13.6 g, 0.14 mol L–1c)1.9 g, 0.28 mol L–1d)13.6 g, 0.28 mol L–1Correct answer is option 'C'. Can you explain this answer?
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A mixture of 100 m mol of Ca(OH)2 and 2g 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; Kspof Ca(OH)2 is 5.5 × 10–6)a)1.9 g, 0.14 mol L–1b)13.6 g, 0.14 mol L–1c)1.9 g, 0.28 mol L–1d)13.6 g, 0.28 mol L–1Correct answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A mixture of 100 m mol of Ca(OH)2 and 2g 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; Kspof Ca(OH)2 is 5.5 × 10–6)a)1.9 g, 0.14 mol L–1b)13.6 g, 0.14 mol L–1c)1.9 g, 0.28 mol L–1d)13.6 g, 0.28 mol L–1Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A mixture of 100 m mol of Ca(OH)2 and 2g 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; Kspof Ca(OH)2 is 5.5 × 10–6)a)1.9 g, 0.14 mol L–1b)13.6 g, 0.14 mol L–1c)1.9 g, 0.28 mol L–1d)13.6 g, 0.28 mol L–1Correct answer is option 'C'. Can you explain this answer?.
Solutions for A mixture of 100 m mol of Ca(OH)2 and 2g 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; Kspof Ca(OH)2 is 5.5 × 10–6)a)1.9 g, 0.14 mol L–1b)13.6 g, 0.14 mol L–1c)1.9 g, 0.28 mol L–1d)13.6 g, 0.28 mol L–1Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
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