To solve the problem of how much `Ca(NO₃)₂` is needed in mg for a solution with 2.35 ppm of Ca in 50 ml, we can follow these steps:
### Step 1: Understand the ppm concentration
The term "ppm" stands for "parts per million." A concentration of 2.35 ppm means there are 2.35 grams of calcium (Ca) in 1,000,000 grams of solution.
### Step 2: Calculate the amount of Ca in 50 ml of solution
Since 1 ml of water weighs approximately 1 gram, 50 ml of solution weighs about 50 grams. We can set up a proportion to find out how much calcium is in 50 ml of solution:
\[
\text{Amount of Ca in 50 ml} = \frac{2.35 \, \text{g Ca}}{10^6 \, \text{g solution}} \times 50 \, \text{g solution}
\]
Calculating this gives:
\[
\text{Amount of Ca} = \frac{2.35 \times 50}{10^6} = 0.1175 \, \text{g}
\]
### Step 3: Convert grams of Ca to moles of Ca
Next, we need to convert the grams of calcium to moles. The molar mass of calcium (Ca) is approximately 40 g/mol.
\[
\text{Moles of Ca} = \frac{0.1175 \, \text{g}}{40 \, \text{g/mol}} = 0.0029375 \, \text{mol}
\]
### Step 4: Relate moles of Ca to moles of `Ca(NO₃)₂`
From the chemical formula, we know that 1 mole of `Ca(NO₃)₂` contains 1 mole of Ca. Therefore, the moles of `Ca(NO₃)₂` required will be the same as the moles of Ca:
\[
\text{Moles of } Ca(NO₃)₂ = 0.0029375 \, \text{mol}
\]
### Step 5: Calculate the mass of `Ca(NO₃)₂`
Now, we need to find the molar mass of `Ca(NO₃)₂`. The molar mass is calculated as follows:
- Calcium (Ca): 40 g/mol
- Nitrogen (N): 14 g/mol (there are 2 N)
- Oxygen (O): 16 g/mol (there are 6 O)
Calculating the total molar mass:
\[
\text{Molar mass of } Ca(NO₃)₂ = 40 + (2 \times 14) + (6 \times 16) = 40 + 28 + 96 = 164 \, \text{g/mol}
\]
Now we can find the mass of `Ca(NO₃)₂`:
\[
\text{Mass of } Ca(NO₃)₂ = \text{Moles} \times \text{Molar mass} = 0.0029375 \, \text{mol} \times 164 \, \text{g/mol} = 0.4825 \, \text{g}
\]
### Step 6: Convert grams to milligrams
Finally, we convert grams to milligrams:
\[
0.4825 \, \text{g} = 482.5 \, \text{mg}
\]
### Conclusion
Thus, the amount of `Ca(NO₃)₂` required in 50 ml of the solution with 2.35 ppm of Ca is approximately **482.5 mg**.
