Two samples of `HCl` of `1.0 M` and `0.25 M` are mixed. Find volumes of these samples taken in order to prepare `0.75 M HCl` solution. Assume no water is added.
(I) `20 mL, 10 mL` (II) `100 mL, 50 mL`k
(III) `40 mL, 20 mL` (IV) `50 mL, 25 mL`

Options

A. I,II,IV
B. I,II
C. II,III,IV
D. I,II,III,IV

Explanation

To solve the problem of mixing two samples of HCl of different molarities to achieve a desired molarity, we can use the concept of molarity and the relationship between the volumes and concentrations of the solutions.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Molarity of first solution (M1) = 1.0 M
- Molarity of second solution (M2) = 0.25 M
- Desired molarity (M3) = 0.75 M

2. **Define Variables:**
- Let V1 = volume of the first solution (1.0 M)
- Let V2 = volume of the second solution (0.25 M)

3. **Use the Molarity Equation:**
The equation relating the molarity and volume of the solutions is:
\[ M1 \cdot V1 + M2 \cdot V2 = M3 \cdot V3 \]
where \( V3 = V1 + V2 \).

4. **Substitute Known Values:**
Substitute the known values into the equation:
\[ 1.0 \cdot V1 + 0.25 \cdot V2 = 0.75 \cdot (V1 + V2) \]

5. **Expand and Rearrange the Equation:**
Expanding the right side gives:
\[ 1.0 \cdot V1 + 0.25 \cdot V2 = 0.75V1 + 0.75V2 \]
Rearranging this gives:
\[ 1.0V1 - 0.75V1 = 0.75V2 - 0.25V2 \]
Simplifying further:
\[ 0.25V1 = 0.5V2 \]

6. **Find the Ratio of Volumes:**
Dividing both sides by 0.25 gives:
\[ V1 = 2V2 \]
This means the ratio of \( V1 : V2 = 2 : 1 \).

7. **Check the Given Options:**
Now we can check the provided options to see which pairs of volumes maintain this ratio:
- (I) 20 mL, 10 mL → \( 20 : 10 = 2 : 1 \) (Valid)
- (II) 100 mL, 50 mL → \( 100 : 50 = 2 : 1 \) (Valid)
- (III) 40 mL, 20 mL → \( 40 : 20 = 2 : 1 \) (Valid)
- (IV) 50 mL, 25 mL → \( 50 : 25 = 2 : 1 \) (Valid)

8. **Conclusion:**
All given options maintain the ratio of 2:1, so they can all be used to prepare a 0.75 M HCl solution.