The volume of water added to 500 mL , 0.5 M NaOH so that its strenth becomes 10mg NaOH per mL is

Options

A. 250 mL
B. 500 mL
C. 750 mL
D. 1000 mL

Answer

To solve the problem of determining the volume of water added to 500 mL of 0.5 M NaOH so that its strength becomes 10 mg NaOH per mL, we can follow these steps:

### Step 1: Convert the desired strength from mg/mL to molarity (M)
1. The desired strength is 10 mg NaOH per mL.
2. Convert mg to grams:
\[ 10 \text{ mg} = 10 \times 10^{-3} \text{ g} = 0.01 \text{ g} \]
3. Calculate the molar mass of NaOH:
\[ \text{Molar mass of NaOH} = 23 \text{ (Na)} + 16 \text{ (O)} + 1 \text{ (H)} = 40 \text{ g/mol} \]
4. Convert grams to moles:
\[ \text{Number of moles} = \frac{0.01 \text{ g}}{40 \text{ g/mol}} = 0.00025 \text{ moles} \]
5. Since this is for 1 mL, the molarity is:
\[ \text{Molarity} = \frac{0.00025 \text{ moles}}{0.001 \text{ L}} = 0.25 \text{ M} \]

### Step 2: Use the dilution formula to find the final volume
1. We can use the dilution equation \( M_1V_1 = M_2V_2 \), where:
- \( M_1 = 0.5 \) M (initial concentration)
- \( V_1 = 500 \) mL (initial volume)
- \( M_2 = 0.25 \) M (final concentration)
- \( V_2 \) is the final volume we need to find.

2. Rearranging the formula to find \( V_2 \):
\[ V_2 = \frac{M_1 \times V_1}{M_2} = \frac{0.5 \times 500}{0.25} \]

### Step 3: Calculate \( V_2 \)
1. Calculate \( V_2 \):
\[ V_2 = \frac{250}{0.25} = 1000 \text{ mL} \]

### Step 4: Determine the volume of water added
1. The volume of water added is:
\[ \text{Volume of water added} = V_2 - V_1 = 1000 \text{ mL} - 500 \text{ mL} = 500 \text{ mL} \]

### Final Answer
The volume of water added to the solution is **500 mL**.

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