Select correct statement regarding a sample of water having 100 ppm of `CaCl_(2)` .

Options

A. `Ca^(+2) = 100` pm, `Cl^(-) = 200` pm
B. `Ca^(+2) = 100` pm, `Cl^(-) = 100` pm
C. `Ca^(+2) = 33` pm, `Cl^(-) = 67` pm
D. `Ca^(+2) = 36` pm, `Cl^(-) = 64` pm

Explanation

To solve the question regarding a sample of water having 100 ppm of \( \text{CaCl}_2 \), we can follow these steps:

### Step 1: Understand the Definition of ppm
Parts per million (ppm) is defined as the mass of the substance divided by the mass of the solution, multiplied by \( 10^6 \):
\[ \text{ppm} = \left( \frac{\text{mass of substance}}{\text{mass of solution}} \right) \times 10^6 \]

### Step 2: Assume the Mass of the Solution
For convenience, we can assume the mass of the solution is \( 10^6 \) grams (1,000,000 grams). This simplifies our calculations.

### Step 3: Calculate the Mass of \( \text{CaCl}_2 \)
Given that the ppm of \( \text{CaCl}_2 \) is 100, we can set up the equation:
\[ 100 = \left( \frac{\text{mass of } \text{CaCl}_2}{10^6} \right) \times 10^6 \]
From this, we can deduce that:
\[ \text{mass of } \text{CaCl}_2 = 100 \text{ grams} \]

### Step 4: Calculate the Moles of \( \text{CaCl}_2 \)
To find the number of moles of \( \text{CaCl}_2 \), we use the formula:
\[ \text{moles} = \frac{\text{mass}}{\text{molar mass}} \]
The molar mass of \( \text{CaCl}_2 \) is calculated as follows:
- Calcium (Ca) = 40 g/mol
- Chlorine (Cl) = 35.5 g/mol (and there are 2 Cl atoms)

Thus, the molar mass of \( \text{CaCl}_2 \) is:
\[ \text{Molar mass of } \text{CaCl}_2 = 40 + (2 \times 35.5) = 40 + 71 = 111 \text{ g/mol} \]

Now, we can calculate the moles:
\[ \text{moles of } \text{CaCl}_2 = \frac{100 \text{ g}}{111 \text{ g/mol}} \approx 0.900 \text{ moles} \]

### Step 5: Determine the Moles of Ions
From the dissociation of \( \text{CaCl}_2 \):
\[ \text{CaCl}_2 \rightarrow \text{Ca}^{2+} + 2 \text{Cl}^- \]
This means:
- 1 mole of \( \text{CaCl}_2 \) produces 1 mole of \( \text{Ca}^{2+} \)
- 1 mole of \( \text{CaCl}_2 \) produces 2 moles of \( \text{Cl}^- \)

Thus, for 0.900 moles of \( \text{CaCl}_2 \):
- Moles of \( \text{Ca}^{2+} = 0.900 \)
- Moles of \( \text{Cl}^- = 2 \times 0.900 = 1.800 \)

### Step 6: Calculate the Mass of Ions
To find the mass of each ion:
- Mass of \( \text{Ca}^{2+} \):
\[ \text{mass of } \text{Ca}^{2+} = \text{moles} \times \text{molar mass} = 0.900 \times 40 \approx 36 \text{ grams} \]
- Mass of \( \text{Cl}^- \):
\[ \text{mass of } \text{Cl}^- = \text{moles} \times \text{molar mass} = 1.800 \times 35.5 \approx 63.9 \text{ grams} \]

### Step 7: Calculate ppm for Each Ion
Using the mass of the solution assumed to be \( 10^6 \) grams:
- ppm of \( \text{Ca}^{2+} \):
\[ \text{ppm of } \text{Ca}^{2+} = \left( \frac{36 \text{ g}}{10^6 \text{ g}} \right) \times 10^6 = 36 \text{ ppm} \]
- ppm of \( \text{Cl}^- \):
\[ \text{ppm of } \text{Cl}^- = \left( \frac{63.9 \text{ g}}{10^6 \text{ g}} \right) \times 10^6 \approx 64 \text{ ppm} \]

### Conclusion
The correct statements regarding the sample of water having 100 ppm of \( \text{CaCl}_2 \) are:
- The concentration of \( \text{Ca}^{2+} \) is approximately 36 ppm.
- The concentration of \( \text{Cl}^- \) is approximately 64 ppm.