250 ml of a calcium carbonate solution contains 2.5 grams of `CaCO_(3)`. If 10 ml of this solution is diluted to one litre, what is the concentration of the resultant solution ?

Options

A. 0.1 M
B. 0.001 M
C. 0.01 M
D. `10^(-4)M`

Explanation

To solve the problem step by step, we will determine the concentration of the resultant solution after diluting a portion of the original calcium carbonate solution.

### Step 1: Calculate the Molarity of the Original Solution

1. **Determine the number of moles of CaCO3**:
- Given mass of CaCO3 = 2.5 grams
- Molar mass of CaCO3 = 40 (Ca) + 12 (C) + 3 × 16 (O) = 100 g/mol
- Number of moles = mass / molar mass = 2.5 g / 100 g/mol = 0.025 moles

2. **Convert the volume of the solution from mL to L**:
- Volume of solution = 250 mL = 250 / 1000 = 0.250 L

3. **Calculate the molarity (M)**:
- Molarity (M) = number of moles / volume in L = 0.025 moles / 0.250 L = 0.1 M

### Step 2: Dilute the Solution

1. **Take 10 mL of the original solution**:
- Volume taken (V1) = 10 mL = 10 / 1000 = 0.010 L

2. **Calculate the number of moles in the 10 mL of the original solution**:
- Moles in 10 mL = Molarity × Volume = 0.1 M × 0.010 L = 0.001 moles

3. **Dilute this 10 mL to 1 L**:
- Final volume (V2) = 1 L

### Step 3: Calculate the Concentration of the Resultant Solution

1. **Use the dilution formula**:
- M1V1 = M2V2
- Where:
- M1 = initial molarity = 0.1 M
- V1 = initial volume = 0.010 L
- M2 = final molarity (to be calculated)
- V2 = final volume = 1 L

2. **Rearranging the formula to find M2**:
- M2 = (M1 × V1) / V2
- M2 = (0.1 M × 0.010 L) / 1 L = 0.001 M

### Final Answer:
The concentration of the resultant solution after dilution is **0.001 M**.

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