A mixture of 200 mmol of `Ca(OH)_2 ` and 3g of sodium sulphate was dissolved in water and the volume was made up to 200 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 xx 10^(-6) ` )

Options

A. `2.9 gm 0.21 mol L^(-1)`
B. `13.6 gm , 0.21 mol L^(-1)`
C. `2.9 gm , 0.11 mol L^(-1)`
D. `13.6 gm , 0.11 molL^(-1)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the number of moles of sodium sulfate (Na₂SO₄)
Given:
- Mass of Na₂SO₄ = 3 g
- Molar mass of Na₂SO₄ = 143 g/mol

Using the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
\[ \text{Number of moles of Na₂SO₄} = \frac{3 \, \text{g}}{143 \, \text{g/mol}} \approx 0.02097 \, \text{mol} \approx 20.97 \, \text{mmol} \]

### Step 2: Identify the limiting reagent
We have:
- Moles of Ca(OH)₂ = 200 mmol
- Moles of Na₂SO₄ = 20.97 mmol

From the reaction:
\[ \text{Ca(OH)}_2 + \text{Na}_2\text{SO}_4 \rightarrow \text{CaSO}_4 + 2 \text{NaOH} \]
The stoichiometry shows that 1 mole of Na₂SO₄ reacts with 1 mole of Ca(OH)₂.

To find the limiting reagent, we compare the available moles:
- For Na₂SO₄: 20.97 mmol
- For Ca(OH)₂: 200 mmol

Since Na₂SO₄ is present in a smaller amount, it is the limiting reagent.

### Step 3: Calculate the mass of calcium sulfate (CaSO₄) formed
From the stoichiometry of the reaction:
1 mole of Na₂SO₄ produces 1 mole of CaSO₄.

Thus, moles of CaSO₄ produced = moles of Na₂SO₄ = 20.97 mmol.

Now, calculate the mass of CaSO₄:
- Molar mass of CaSO₄ = 136 g/mol

Using the formula:
\[ \text{Mass} = \text{Number of moles} \times \text{Molar mass} \]
\[ \text{Mass of CaSO₄} = 20.97 \, \text{mmol} \times \frac{136 \, \text{g/mol}}{1000} \approx 2.856 \, \text{g} \approx 2.9 \, \text{g} \]

### Step 4: Calculate the concentration of OH⁻ ions in the solution
From the reaction:
1 mole of Ca(OH)₂ produces 2 moles of NaOH, which means it produces 2 moles of OH⁻ ions.

Thus, moles of OH⁻ produced = 2 × moles of Ca(OH)₂ reacted.

Since Na₂SO₄ is the limiting reagent and it reacts with Ca(OH)₂, we can find the moles of Ca(OH)₂ that reacted:
- Moles of Ca(OH)₂ that reacted = 20.97 mmol (same as Na₂SO₄)

Thus, moles of OH⁻ produced:
\[ \text{Moles of OH}^- = 2 \times 20.97 \, \text{mmol} = 41.94 \, \text{mmol} \]

Now, calculate the concentration of OH⁻:
\[ \text{Concentration of OH}^- = \frac{\text{Number of moles}}{\text{Volume in L}} = \frac{41.94 \, \text{mmol}}{200 \, \text{mL}} = \frac{41.94 \, \text{mmol}}{0.2 \, \text{L}} = 0.2097 \, \text{mol/L} \approx 0.21 \, \text{mol/L} \]

### Final Results
- Mass of CaSO₄ formed: **2.9 g**
- Concentration of OH⁻ in the solution: **0.21 mol/L**