If water contains `10 p p m` of `MgCl_(2)` and `8p p m of CaSO_(4)` calculate the `p p m of CaCO_(3)`

Explanation

To calculate the parts per million (ppm) of CaCO₃ in water containing 10 ppm of MgCl₂ and 8 ppm of CaSO₄, we can follow these steps:

### Step 1: Understand the Given Data
We are given:
- 10 ppm of MgCl₂
- 8 ppm of CaSO₄

### Step 2: Calculate the Mass of CaSO₄ in Water
Since ppm is defined as parts per million, this means:
- 8 ppm of CaSO₄ means that in 1,000,000 grams (or 1,000 liters) of water, there are 8 grams of CaSO₄.

### Step 3: Calculate the Moles of CaSO₄
To find the moles of CaSO₄, we need its molar mass. The molar mass of CaSO₄ is calculated as follows:
- Calcium (Ca) = 40 g/mol
- Sulfur (S) = 32 g/mol
- Oxygen (O) = 16 g/mol × 4 = 64 g/mol
- Total molar mass of CaSO₄ = 40 + 32 + 64 = 136 g/mol

Now, we can calculate the moles of CaSO₄:
\[ \text{Moles of CaSO₄} = \frac{\text{Mass of CaSO₄}}{\text{Molar mass of CaSO₄}} = \frac{8 \text{ g}}{136 \text{ g/mol}} = \frac{8}{136} \approx 0.0588 \text{ mol} \]

### Step 4: Determine the Moles of Calcium (Ca)
From the formula of CaSO₄, we see that 1 mole of CaSO₄ contains 1 mole of Ca. Therefore, the moles of Ca will also be:
\[ \text{Moles of Ca} = 0.0588 \text{ mol} \]

### Step 5: Calculate the Moles of CaCO₃
Since each mole of Ca can form one mole of CaCO₃, the moles of CaCO₃ will also be:
\[ \text{Moles of CaCO₃} = 0.0588 \text{ mol} \]

### Step 6: Calculate the Mass of CaCO₃
The molar mass of CaCO₃ is calculated as follows:
- Calcium (Ca) = 40 g/mol
- Carbon (C) = 12 g/mol
- Oxygen (O) = 16 g/mol × 3 = 48 g/mol
- Total molar mass of CaCO₃ = 40 + 12 + 48 = 100 g/mol

Now, we can calculate the mass of CaCO₃:
\[ \text{Mass of CaCO₃} = \text{Moles of CaCO₃} \times \text{Molar mass of CaCO₃} = 0.0588 \text{ mol} \times 100 \text{ g/mol} = 5.88 \text{ g} \]

### Step 7: Calculate the ppm of CaCO₃
Since we have calculated that in 1,000,000 grams of water, there are 5.88 grams of CaCO₃, we can express this as:
\[ \text{ppm of CaCO₃} = \frac{\text{Mass of CaCO₃}}{\text{Mass of water}} \times 10^6 = \frac{5.88 \text{ g}}{1,000,000 \text{ g}} \times 10^6 = 5.88 \text{ ppm} \]

### Final Answer
The parts per million (ppm) of CaCO₃ in the water is **5.88 ppm**.

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