Find the molaity of water. Given:
`rho =1000kg//m^(3)`
[Report your answer upto one decimal place].

Answer

To find the molality of water given its density, we can follow these steps:

### Step 1: Understand the given density
The density of water is given as \( \rho = 1000 \, \text{kg/m}^3 \). This means that 1 cubic meter of water has a mass of 1000 kg.

### Step 2: Convert density to a more useful unit
Since molality is defined as moles of solute per kilogram of solvent, we can convert the density from kg/m³ to kg/L. We know that:
- \( 1 \, \text{m}^3 = 1000 \, \text{L} \)

Thus, the density in kg/L is:
\[ \rho = \frac{1000 \, \text{kg}}{1000 \, \text{L}} = 1 \, \text{kg/L} \]

### Step 3: Calculate the number of moles of water in 1 kg
The molar mass of water (H₂O) is approximately 18 g/mol. To find the number of moles in 1 kg (or 1000 g) of water:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} = \frac{1000 \, \text{g}}{18 \, \text{g/mol}} \approx 55.56 \, \text{moles} \]

### Step 4: Calculate the molality
Molality (m) is defined as the number of moles of solute per kilogram of solvent. In this case, since we are considering pure water, the solvent is also water. Therefore, the molality of water is:
\[ \text{Molality} = \frac{\text{Number of moles}}{\text{Mass of solvent (kg)}} = \frac{55.56 \, \text{moles}}{1 \, \text{kg}} = 55.56 \, \text{mol/kg} \]

### Step 5: Round to one decimal place
Finally, rounding 55.56 to one decimal place gives us:
\[ \text{Molality} \approx 55.6 \, \text{mol/kg} \]

### Final Answer:
The molality of water is **55.6 mol/kg**.

---