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

Answer

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

### Step 1: Understand the given concentration
We have a 10^(-3) molar solution of MgSO4. This means there are 10^(-3) moles of MgSO4 in 1 liter of solution.

### Step 2: Relate moles of MgSO4 to moles of CaCO3
In water, MgSO4 contributes to hardness similarly to CaCO3. Therefore, we can assume that the moles of CaCO3 produced will be equal to the moles of MgSO4 present in the solution.

\[ \text{Moles of CaCO3} = \text{Moles of MgSO4} = 10^{-3} \text{ moles} \]

### Step 3: Calculate the mass of CaCO3
To find the mass of CaCO3, we multiply the number of moles by the molar mass of CaCO3. The molar mass of CaCO3 is approximately 100 g/mol.

\[ \text{Mass of CaCO3} = \text{Moles of CaCO3} \times \text{Molar Mass of CaCO3} = 10^{-3} \text{ moles} \times 100 \text{ g/mol} = 0.1 \text{ g} \]

### Step 4: Calculate the mass of the solution
Assuming the density of the solution is approximately 1 g/mL, the mass of 1 liter (1000 mL) of the solution is:

\[ \text{Mass of solution} = 1000 \text{ mL} \times 1 \text{ g/mL} = 1000 \text{ g} \]

### Step 5: Calculate the degree of hardness in ppm
The degree of hardness in ppm (parts per million) is calculated using the formula:

\[ \text{Degree of hardness (ppm)} = \frac{\text{Mass of CaCO3}}{\text{Mass of solution}} \times 10^6 \]

Substituting the values we found:

\[ \text{Degree of hardness (ppm)} = \frac{0.1 \text{ g}}{1000 \text{ g}} \times 10^6 = 100 \text{ ppm} \]

### Final Answer
Thus, the degree of hardness in terms of ppm of CaCO3 for the 10^(-3) molar MgSO4 solution is **100 ppm**.

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