What mass of water (at `4.0^(@)`C) can be contained in a rectangular box whose dimensions are 10.0 cm by 5.00 cm by 1.00 cm? The density of water at `4.0^(@)`C is `1000 kg//m^(3)`.

Options

A. 5.0 g
B. 10.0 g
C. 25.0 g
D. 50.0 g

Answer

To find the mass of water that can be contained in a rectangular box with given dimensions, we can follow these steps:

### Step 1: Calculate the Volume of the Box
The volume \( V \) of a rectangular box can be calculated using the formula:
\[ V = \text{length} \times \text{width} \times \text{height} \]
Given the dimensions of the box:
- Length = 10.0 cm
- Width = 5.0 cm
- Height = 1.0 cm

Substituting the values:
\[ V = 10.0 \, \text{cm} \times 5.0 \, \text{cm} \times 1.0 \, \text{cm} = 50.0 \, \text{cm}^3 \]

### Step 2: Convert Volume to Cubic Meters
Since the density given is in kg/m³, we need to convert the volume from cm³ to m³. The conversion factor is:
\[ 1 \, \text{cm}^3 = 1 \times 10^{-6} \, \text{m}^3 \]
Thus,
\[ V = 50.0 \, \text{cm}^3 = 50.0 \times 10^{-6} \, \text{m}^3 = 5.0 \times 10^{-5} \, \text{m}^3 \]

### Step 3: Use the Density to Find the Mass
The mass \( m \) can be calculated using the formula:
\[ m = \rho \times V \]
where \( \rho \) is the density of water at \( 4.0^\circ C \), which is given as \( 1000 \, \text{kg/m}^3 \).

Substituting the values:
\[ m = 1000 \, \text{kg/m}^3 \times 5.0 \times 10^{-5} \, \text{m}^3 \]
\[ m = 0.0500 \, \text{kg} \]

### Step 4: Convert Mass to Grams
Since \( 1 \, \text{kg} = 1000 \, \text{g} \):
\[ m = 0.0500 \, \text{kg} \times 1000 \, \text{g/kg} = 50.0 \, \text{g} \]

### Conclusion
The mass of water that can be contained in the box is:
\[ \boxed{50.0 \, \text{g}} \]