5 gm of `K_2 SO_4` solution were dissolved in 250 ml of water. How many ml of this solution are needed to precipitate out 1.2 gm of `BaSO_4` on addition of `BaCI_2`?

Options

A. 44.8 ml
B. 22.4 ml
C. 11.8 ml
D. 22400 ml

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Write the reaction
The reaction between barium chloride (BaCl₂) and potassium sulfate (K₂SO₄) to form barium sulfate (BaSO₄) is as follows:
\[ \text{BaCl}_2 + \text{K}_2\text{SO}_4 \rightarrow \text{BaSO}_4 \downarrow + 2\text{KCl} \]

### Step 2: Calculate the moles of BaSO₄
We need to find out how many moles of BaSO₄ correspond to 1.2 grams. The molar mass of BaSO₄ is calculated as follows:
- Atomic mass of Ba = 137 g/mol
- Atomic mass of S = 32 g/mol
- Atomic mass of O = 16 g/mol (4 O atoms)

So, the molar mass of BaSO₄ is:
\[ 137 + 32 + (4 \times 16) = 137 + 32 + 64 = 233 \text{ g/mol} \]

Now, we can calculate the moles of BaSO₄:
\[ \text{Moles of BaSO}_4 = \frac{\text{mass}}{\text{molar mass}} = \frac{1.2 \text{ g}}{233 \text{ g/mol}} \approx 0.00515 \text{ moles} \]

### Step 3: Relate moles of K₂SO₄ to moles of BaSO₄
From the balanced chemical equation, we can see that the stoichiometry between K₂SO₄ and BaSO₄ is 1:1. Therefore, the moles of K₂SO₄ required will also be 0.00515 moles.

### Step 4: Calculate the mass of K₂SO₄ required
Next, we need to calculate the mass of K₂SO₄ that corresponds to 0.00515 moles. The molar mass of K₂SO₄ is calculated as follows:
- Atomic mass of K = 39 g/mol (2 K atoms)
- Atomic mass of S = 32 g/mol
- Atomic mass of O = 16 g/mol (4 O atoms)

So, the molar mass of K₂SO₄ is:
\[ (2 \times 39) + 32 + (4 \times 16) = 78 + 32 + 64 = 174 \text{ g/mol} \]

Now, we can calculate the mass of K₂SO₄:
\[ \text{Mass of K}_2\text{SO}_4 = \text{moles} \times \text{molar mass} = 0.00515 \text{ moles} \times 174 \text{ g/mol} \approx 0.8961 \text{ g} \]

### Step 5: Determine the concentration of K₂SO₄ in the solution
We know that 5 grams of K₂SO₄ are dissolved in 250 mL of solution. To find the concentration:
\[ \text{Concentration} = \frac{\text{mass}}{\text{volume}} = \frac{5 \text{ g}}{250 \text{ mL}} = 0.02 \text{ g/mL} \]

### Step 6: Calculate the volume of the solution needed for 0.8961 g of K₂SO₄
Now, we need to find out how much volume of this solution contains 0.8961 g of K₂SO₄:
\[ \text{Volume} = \frac{\text{mass}}{\text{concentration}} = \frac{0.8961 \text{ g}}{0.02 \text{ g/mL}} = 44.805 \text{ mL} \]

### Final Answer
Thus, the volume of the K₂SO₄ solution needed to precipitate out 1.2 g of BaSO₄ is approximately **44.8 mL**.

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