The concentration of `Ca (HCO_(3))_(2)` in a sample of hard water is 405 ppm. The denstiy of water sample is 1.0 gm/ml. Calculate the
molarity of solution ?

Answer

To calculate the molarity of the solution of `Ca(HCO₃)₂` in hard water with a concentration of 405 ppm, we can follow these steps:

### Step 1: Understand the definition of ppm
PPM (parts per million) is defined as the mass of solute per million parts of the solution. In this case, 405 ppm means there are 405 grams of `Ca(HCO₃)₂` in 1,000,000 grams of the solution.

### Step 2: Convert ppm to grams of solute
Since the density of the water sample is given as 1.0 g/ml, we can assume that 1 liter (1000 ml) of the solution has a mass of 1000 grams. Therefore, we can calculate the mass of the solute in 1 liter of solution:
\[ \text{Mass of solute in 1 liter} = \frac{405 \text{ grams}}{10^6 \text{ grams}} \times 1000 \text{ grams} = 0.405 \text{ grams} \]

### Step 3: Calculate the molar mass of `Ca(HCO₃)₂`
To find the molarity, we need to calculate the number of moles of `Ca(HCO₃)₂`. The molar mass can be calculated as follows:
- Calcium (Ca): 40.08 g/mol
- Hydrogen (H): 1.01 g/mol (2 H in `Ca(HCO₃)₂`)
- Carbon (C): 12.01 g/mol (1 C in `Ca(HCO₃)₂`, and there are 2 HCO₃ groups)
- Oxygen (O): 16.00 g/mol (3 O in `HCO₃`, and there are 2 HCO₃ groups)

Calculating the molar mass:
\[ \text{Molar mass} = 40.08 + (2 \times 1.01) + (2 \times 12.01) + (6 \times 16.00) \]
\[ = 40.08 + 2.02 + 24.02 + 96.00 = 162.12 \text{ g/mol} \]

### Step 4: Calculate the number of moles of `Ca(HCO₃)₂`
Now, we can calculate the number of moles of `Ca(HCO₃)₂` in 0.405 grams:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.405 \text{ grams}}{162.12 \text{ g/mol}} \approx 0.0025 \text{ moles} \]

### Step 5: Calculate the molarity of the solution
Molarity (M) is defined as the number of moles of solute per liter of solution. Since we have calculated the number of moles and we are considering 1 liter of solution:
\[ \text{Molarity} = \frac{\text{number of moles}}{\text{volume of solution in liters}} = \frac{0.0025 \text{ moles}}{1 \text{ liter}} = 0.0025 \text{ M} \]

### Final Answer
The molarity of the `Ca(HCO₃)₂` solution is approximately **0.0025 M**.

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