Determine degree of hardness in term of ppm of `CaCO_3` of `10^(-3)` molar `MgSO_4` (aq).

Options

A. (a) 100 ppm
B. (b) 200 ppm
C. (c) 300 ppm
D. (d) 400 ppm

Explanation

To determine the degree of hardness in terms of ppm of CaCO3 from a 10^(-3) molar MgSO4 solution, we can follow these steps:

### Step 1: Understand the relationship between MgSO4 and CaCO3
1. **Given**: 10^(-3) molar MgSO4 means there are 10^(-3) moles of MgSO4 in 1 liter of solution.
2. **Note**: Each mole of MgSO4 contributes to the hardness of water due to the presence of Mg²⁺ ions, which can be expressed in terms of CaCO3.

### Step 2: Calculate the moles of CaCO3
1. **Molarity of CaCO3**: Since 1 mole of MgSO4 can be considered equivalent to 1 mole of CaCO3 in terms of hardness, we have:
\[ \text{Moles of CaCO3} = 10^{-3} \text{ moles} \]

### Step 3: Calculate the mass of CaCO3
1. **Molar mass of CaCO3**: The molar mass of CaCO3 is approximately 100 g/mol.
2. **Mass of CaCO3**: To find the mass of CaCO3 in grams:
\[ \text{Mass of CaCO3} = \text{Moles} \times \text{Molar mass} = 10^{-3} \text{ moles} \times 100 \text{ g/mol} = 0.1 \text{ g} \]

### Step 4: Convert mass to ppm
1. **Understanding ppm**: PPM (parts per million) is defined as:
\[ \text{ppm} = \left(\frac{\text{mass of solute (g)}}{\text{mass of solution (g)}}\right) \times 10^6 \]
2. **Assuming the density of water**: For 1 liter of solution, the mass of the solution is approximately 1000 g (since the density of water is about 1 g/cm³).
3. **Calculating ppm**:
\[ \text{ppm} = \left(\frac{0.1 \text{ g}}{1000 \text{ g}}\right) \times 10^6 = 100 \text{ ppm} \]

### Conclusion
The degree of hardness in terms of ppm of CaCO3 for a 10^(-3) molar MgSO4 solution is **100 ppm**.

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