How many millilitree of `6.0` M hydrochloric acid should be used to prepare 150 ml of 0.30 M HCl solution ?

Options

A. `3.0`
B. `7.5`
C. `9.3`
D. `30`

Answer

To solve the problem of how many milliliters of 6.0 M hydrochloric acid (HCl) are needed to prepare 150 mL of a 0.30 M HCl solution, we can use the concept of molarity and the dilution equation.

### Step-by-Step Solution:

1. **Understand the Molarity Concept**:
Molarity (M) is defined as the number of moles of solute per liter of solution. The formula for calculating the number of moles is:
\[ \text{Number of moles} = \text{Molarity} \times \text{Volume (in liters)} \]

2. **Identify Given Values**:
- Molarity of concentrated HCl (M1) = 6.0 M
- Molarity of diluted HCl (M2) = 0.30 M
- Volume of diluted HCl (V2) = 150 mL = 0.150 L (convert mL to L)

3. **Use the Dilution Equation**:
The dilution equation relates the molarity and volume of the concentrated and diluted solutions:
\[ M1 \times V1 = M2 \times V2 \]
where:
- \( V1 \) is the volume of the concentrated solution we need to find.

4. **Substitute Known Values**:
Plugging in the known values into the equation:
\[ 6.0 \, \text{M} \times V1 = 0.30 \, \text{M} \times 0.150 \, \text{L} \]

5. **Calculate the Right Side**:
Calculate the right side of the equation:
\[ 0.30 \, \text{M} \times 0.150 \, \text{L} = 0.045 \, \text{moles} \]

6. **Rearrange to Solve for \( V1 \)**:
Rearranging the equation to solve for \( V1 \):
\[ V1 = \frac{0.045 \, \text{moles}}{6.0 \, \text{M}} = 0.0075 \, \text{L} \]

7. **Convert to Milliliters**:
Convert \( V1 \) from liters to milliliters:
\[ V1 = 0.0075 \, \text{L} \times 1000 \, \text{mL/L} = 7.5 \, \text{mL} \]

### Final Answer:
You will need **7.5 mL** of 6.0 M hydrochloric acid to prepare 150 mL of a 0.30 M HCl solution.