`50mLof `20.8% (w/V) Ba CI_2 and (aq) and 100mL` of `9.8%mL(w/V) H_2SO_4(aq)` solutions are mixed. Molarity of `CI^- iron in the resulting solution is : (At mass of `Ba=137`)

Options

A. 0.333M
B. 0.666M
C. 0.1M
D. 1.33M

Answer

To find the molarity of chloride ions (Cl⁻) in the resulting solution after mixing barium chloride (BaCl₂) and sulfuric acid (H₂SO₄), we can follow these steps:

### Step 1: Calculate the mass of BaCl₂ in the solution
Given that the barium chloride solution is 20.8% (w/V), this means there are 20.8 grams of BaCl₂ in 100 mL of solution.

For 50 mL of this solution:
\[ \text{Mass of BaCl}_2 = \left(\frac{20.8 \, \text{g}}{100 \, \text{mL}}\right) \times 50 \, \text{mL} = 10.4 \, \text{g} \]

### Step 2: Calculate the number of moles of BaCl₂
The molar mass of BaCl₂ can be calculated as follows:
- Atomic mass of Ba = 137 g/mol
- Atomic mass of Cl = 35.5 g/mol
- Molar mass of BaCl₂ = 137 + (2 × 35.5) = 137 + 71 = 208 g/mol

Now, calculate the number of moles of BaCl₂:
\[ \text{Number of moles of BaCl}_2 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{10.4 \, \text{g}}{208 \, \text{g/mol}} = 0.05 \, \text{moles} \]

### Step 3: Calculate the mass of H₂SO₄ in the solution
Given that the H₂SO₄ solution is 9.8% (w/V), this means there are 9.8 grams of H₂SO₄ in 100 mL of solution.

For 100 mL of this solution:
\[ \text{Mass of H}_2\text{SO}_4 = 9.8 \, \text{g} \]

### Step 4: Calculate the number of moles of H₂SO₄
The molar mass of H₂SO₄ can be calculated as follows:
- Atomic mass of H = 1 g/mol
- Atomic mass of S = 32 g/mol
- Atomic mass of O = 16 g/mol
- Molar mass of H₂SO₄ = (2 × 1) + 32 + (4 × 16) = 2 + 32 + 64 = 98 g/mol

Now, calculate the number of moles of H₂SO₄:
\[ \text{Number of moles of H}_2\text{SO}_4 = \frac{9.8 \, \text{g}}{98 \, \text{g/mol}} = 0.1 \, \text{moles} \]

### Step 5: Determine the limiting reagent
From the balanced chemical reaction:
\[ \text{BaCl}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{BaSO}_4 + 2 \text{HCl} \]
1 mole of BaCl₂ reacts with 1 mole of H₂SO₄ to produce 2 moles of HCl.

Since we have 0.05 moles of BaCl₂ and 0.1 moles of H₂SO₄, BaCl₂ is the limiting reagent.

### Step 6: Calculate the moles of HCl produced
From the stoichiometry of the reaction:
\[ 0.05 \, \text{moles of BaCl}_2 \rightarrow 2 \times 0.05 = 0.1 \, \text{moles of HCl} \]

### Step 7: Calculate the moles of Cl⁻ ions produced
Since each mole of HCl produces 1 mole of Cl⁻:
\[ \text{Moles of Cl}^- = 0.1 \, \text{moles} \]

### Step 8: Calculate the total volume of the resulting solution
The total volume of the solution after mixing is:
\[ 50 \, \text{mL} + 100 \, \text{mL} = 150 \, \text{mL} = 0.15 \, \text{L} \]

### Step 9: Calculate the molarity of Cl⁻ ions
Molarity (M) is defined as the number of moles of solute per liter of solution:
\[ \text{Molarity of Cl}^- = \frac{\text{Moles of Cl}^-}{\text{Volume of solution in L}} = \frac{0.1 \, \text{moles}}{0.15 \, \text{L}} = 0.6667 \, \text{M} \]

Thus, the molarity of chloride ions in the resulting solution is approximately **0.666 M**.