25 mL of 3 M HCl were added to 75 mL of 0.05 M HCl. The molarity of HCl in the resulting solution is approximately-

Options

A. 0.055 M
B. 0.35 M
C. 0.787 M
D. 3.05 M

Answer

To find the molarity of HCl in the resulting solution after mixing 25 mL of 3 M HCl with 75 mL of 0.05 M HCl, we can follow these steps:

### Step 1: Calculate the number of moles of HCl in each solution before mixing.

1. **For 25 mL of 3 M HCl:**
\[ \text{Moles of HCl} = \text{Molarity} \times \text{Volume} = 3 \, \text{mol/L} \times 0.025 \, \text{L} = 0.075 \, \text{mol} \]

2. **For 75 mL of 0.05 M HCl:**
\[ \text{Moles of HCl} = \text{Molarity} \times \text{Volume} = 0.05 \, \text{mol/L} \times 0.075 \, \text{L} = 0.00375 \, \text{mol} \]

### Step 2: Calculate the total number of moles of HCl after mixing.

\[ \text{Total moles of HCl} = \text{Moles from 3 M solution} + \text{Moles from 0.05 M solution} \]
\[ \text{Total moles of HCl} = 0.075 \, \text{mol} + 0.00375 \, \text{mol} = 0.07875 \, \text{mol} \]

### Step 3: Calculate the total volume of the resulting solution.

\[ \text{Total volume} = 25 \, \text{mL} + 75 \, \text{mL} = 100 \, \text{mL} = 0.1 \, \text{L} \]

### Step 4: Calculate the molarity of the resulting solution.

\[ \text{Molarity (M)} = \frac{\text{Total moles of HCl}}{\text{Total volume in L}} = \frac{0.07875 \, \text{mol}}{0.1 \, \text{L}} = 0.7875 \, \text{M} \]

### Final Answer:
The molarity of HCl in the resulting solution is approximately **0.79 M**.

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