If hardness of water sample is 200ppm, then select the incorrect statement:

Options

A. `"Mass ratio of"CaCO_(3) "to" H_(2)O is (0.02)/(100)`
B. `"Mole ratio of"CaCO_(3) "to" H_(2)O is 3.6 xx 10^-5`
C. `"Mass of" CaCO_(3) "present in hard water" is 0.2g//L`
D. `"1 miliequivalent of" CaCO_(3) "present in 1kg of hard water"`

Answer

To solve the problem of determining the incorrect statement regarding the hardness of a water sample with a hardness of 200 ppm, we can follow these steps:

### Step 1: Understand the Definition of Hardness in ppm
The hardness of water in ppm (parts per million) is defined as the mass of calcium carbonate (CaCO₃) present in a given mass of water. Specifically, it is calculated as:
\[ \text{Hardness (ppm)} = \frac{\text{mass of CaCO}_3}{\text{mass of water}} \times 10^6 \]
Given that the hardness is 200 ppm, this means:
\[ 200 \text{ g of CaCO}_3 is present in } 10^6 \text{ g of water.} \]

### Step 2: Calculate the Mass Ratio of Calcium Carbonate to Water
To find the mass ratio of calcium carbonate to water:
\[ \text{Mass ratio} = \frac{200 \text{ g}}{10^6 \text{ g}} = 2 \times 10^{-4} \]
This can also be expressed as:
\[ \frac{0.02}{100} \]
Thus, the first option stating that the mass ratio of calcium carbonate to water is \(0.02/100\) is **correct**.

### Step 3: Calculate the Mole Ratio of Calcium Carbonate to Water
Next, we calculate the number of moles of calcium carbonate and water:
- Moles of CaCO₃:
\[ \text{Moles of CaCO}_3 = \frac{200 \text{ g}}{100 \text{ g/mol}} = 2 \text{ moles} \]
- Moles of water (H₂O):
\[ \text{Moles of H}_2\text{O} = \frac{10^6 \text{ g}}{18 \text{ g/mol}} = \frac{10^6}{18} \text{ moles} \]
Now, calculate the mole ratio:
\[ \text{Mole ratio} = \frac{\text{Moles of CaCO}_3}{\text{Moles of H}_2\text{O}} = \frac{2}{\frac{10^6}{18}} = \frac{36}{10^6} = 3.6 \times 10^{-5} \]
Thus, the second option stating that the mole ratio of calcium carbonate to water is \(3.6 \times 10^{-5}\) is **correct**.

### Step 4: Calculate the Mass of Calcium Carbonate per Liter of Water
To find the mass of calcium carbonate in 1 liter of water:
- Since \(10^6 \text{ g of water} = 1000 \text{ liters}\):
\[ \text{Mass of CaCO}_3 \text{ in 1 liter} = \frac{200 \text{ g}}{1000 \text{ liters}} = 0.2 \text{ g/liter} \]
Thus, the third option stating that the mass of calcium carbonate present in hard water is \(0.2 \text{ g/liter}\) is **correct**.

### Step 5: Calculate Milliequivalents of Calcium Carbonate in 1 kg of Water
Now, we need to determine the milliequivalents of calcium carbonate in 1 kg of water:
- Convert 200 g of CaCO₃ to milligrams:
\[ 200 \text{ g} = 200 \times 10^3 \text{ mg} \]
- Calculate the number of millimoles:
\[ \text{Millimoles of CaCO}_3 = \frac{200 \times 10^3 \text{ mg}}{100 \text{ mg/mmol}} = 2000 \text{ mmol} \]
- The n-factor for CaCO₃ (which dissociates into Ca²⁺ and CO₃²⁻) is 2:
\[ \text{Milliequivalents} = \text{Millimoles} \times \text{n-factor} = 2000 \text{ mmol} \times 2 = 4000 \text{ meq} \]
The fourth option states that 1 milliequivalent of calcium carbonate is present in 1 kg of hard water, which is **incorrect**.

### Conclusion
The incorrect statement among the options is the fourth one.

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