Washing soda `(Na_(2)CO_(3).10H_(2)O)` is widely used in softening of hard waer. If 1 L of hard water requires 0.0286 g of washing soda, the hardness of `CaCO_(3)` in ppm is

Options

A. 10 ppm
B. 5 ppm
C. 8 ppm
D. 6 ppm

Explanation

To solve the problem, we need to find the hardness of water in terms of CaCO₃ (calcium carbonate) in parts per million (ppm) based on the amount of washing soda (Na₂CO₃.10H₂O) used. Here is a step-by-step solution:

### Step 1: Calculate the moles of washing soda used
Given that 1 L of hard water requires 0.0286 g of washing soda, we first need to calculate the moles of washing soda.

**Molar mass of washing soda (Na₂CO₃.10H₂O):**
- Sodium (Na): 23 g/mol × 2 = 46 g/mol
- Carbon (C): 12 g/mol × 1 = 12 g/mol
- Oxygen (O): 16 g/mol × 3 = 48 g/mol
- Water (H₂O): 18 g/mol × 10 = 180 g/mol

Total molar mass = 46 + 12 + 48 + 180 = 286 g/mol

Now, calculate the moles of washing soda:
\[ \text{Moles of Na}_2\text{CO}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{0.0286 \text{ g}}{286 \text{ g/mol}} \approx 1.00 \times 10^{-4} \text{ moles} \]

### Step 2: Relate moles of Na₂CO₃ to moles of CaCO₃
From the reaction:
\[ \text{Na}_2\text{CO}_3 + \text{Ca}^{2+} \rightarrow 2\text{Na}^+ + \text{CaCO}_3 \]
1 mole of Na₂CO₃ reacts with 1 mole of Ca²⁺ to produce 1 mole of CaCO₃.

Thus, the moles of CaCO₃ produced will also be:
\[ \text{Moles of CaCO}_3 = 1.00 \times 10^{-4} \text{ moles} \]

### Step 3: Calculate the mass of CaCO₃ produced
Using the molar mass of CaCO₃:
- 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, calculate the mass of CaCO₃:
\[ \text{Mass of CaCO}_3 = \text{moles} \times \text{molar mass} = 1.00 \times 10^{-4} \text{ moles} \times 100 \text{ g/mol} = 0.01 \text{ g} \]

### Step 4: Calculate the hardness in ppm
Parts per million (ppm) is calculated using the formula:
\[ \text{ppm} = \left( \frac{\text{mass of solute (g)}}{\text{mass of solution (g)}} \right) \times 10^6 \]
Assuming the mass of 1 L of water is approximately 1000 g (since the density of water is about 1 g/mL):
\[ \text{ppm} = \left( \frac{0.01 \text{ g}}{1000 \text{ g}} \right) \times 10^6 = 10 \text{ ppm} \]

### Final Answer
The hardness of CaCO₃ in the water is **10 ppm**.

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