A full can of soda has a mass of 0.416 kg. It contains `3.54 xx 10^(-4) m^(3)` of liquid. Assuming that the soda has the same density as water, find the volume of aluminum used to make the can.

Options

A. `6.2 xx 10^(-5) m^(3)`
B. `4.2 xx 10^(-5) m^(3)`
C. `2.3 xx 10^(-5) m^(3)`
D. `1.8 xx 10^(-5) m^(3)`

Answer

To find the volume of aluminum used to make the can, we can follow these steps:

### Step 1: Determine the mass of the liquid (soda) in the can.
Given:
- Total mass of the can (including soda) = 0.416 kg
- Volume of soda = \(3.54 \times 10^{-4} \, m^3\)
- Density of soda (same as water) = \(1000 \, kg/m^3\)

Using the formula for mass:
\[ \text{Mass of liquid (soda)} = \text{Density} \times \text{Volume} \]
\[ \text{Mass of liquid (soda)} = 1000 \, kg/m^3 \times 3.54 \times 10^{-4} \, m^3 = 0.354 \, kg \]

### Step 2: Calculate the mass of aluminum used in the can.
To find the mass of aluminum, we subtract the mass of the soda from the total mass of the can:
\[ \text{Mass of aluminum} = \text{Total mass} - \text{Mass of liquid (soda)} \]
\[ \text{Mass of aluminum} = 0.416 \, kg - 0.354 \, kg = 0.062 \, kg \]

### Step 3: Determine the volume of aluminum used.
We need the density of aluminum to find its volume. The density of aluminum is approximately \(2700 \, kg/m^3\).

Using the formula for volume:
\[ \text{Volume of aluminum} = \frac{\text{Mass of aluminum}}{\text{Density of aluminum}} \]
\[ \text{Volume of aluminum} = \frac{0.062 \, kg}{2700 \, kg/m^3} = 2.296 \times 10^{-5} \, m^3 \]

### Step 4: Round the volume to appropriate significant figures.
The volume of aluminum used to make the can is approximately:
\[ \text{Volume of aluminum} \approx 2.3 \times 10^{-5} \, m^3 \]

### Final Answer:
The volume of aluminum used to make the can is approximately \(2.3 \times 10^{-5} \, m^3\).

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