The active mass of water at `4^(0)C` is

Options

A. 5.55
B. 55.5
C. 0.55
D. Data in sufficient

Explanation

To find the active mass of water at \(4^{\circ}C\), we can follow these steps:

### Step-by-Step Solution:

1. **Understand Active Mass**:
Active mass is defined as the concentration of a substance in a solution, typically expressed in moles per liter (mol/L).

2. **Consider a Sample of Water**:
We will consider a 1-liter sample of water for our calculations.

3. **Density of Water**:
The density of water is approximately \(1 \text{ g/mL}\). This means that 1 liter (which is 1000 mL) of water has a mass of:
\[ \text{Mass of water} = \text{Density} \times \text{Volume} = 1 \text{ g/mL} \times 1000 \text{ mL} = 1000 \text{ g} \]

4. **Calculate the Number of Moles of Water**:
The number of moles can be calculated using the formula:
\[ \text{Number of moles} = \frac{\text{Mass}}{\text{Molar mass}} \]
The molar mass of water (H₂O) is calculated as follows:
- Hydrogen (H) has a molar mass of approximately 1 g/mol, and there are 2 hydrogen atoms in water.
- Oxygen (O) has a molar mass of approximately 16 g/mol.
Thus, the molar mass of water is:
\[ \text{Molar mass of water} = (2 \times 1) + 16 = 18 \text{ g/mol} \]
Now, substituting the values:
\[ \text{Number of moles of water} = \frac{1000 \text{ g}}{18 \text{ g/mol}} \approx 55.56 \text{ moles} \]

5. **Calculate Active Mass**:
The active mass of water is then calculated as:
\[ \text{Active mass} = \frac{\text{Number of moles}}{\text{Volume in liters}} = \frac{55.56 \text{ moles}}{1 \text{ L}} = 55.56 \text{ mol/L} \]

6. **Final Answer**:
Therefore, the active mass of water at \(4^{\circ}C\) is approximately:
\[ \text{Active mass of water} = 55.56 \text{ mol/L} \]