Questions: The mass of a cube of metal is 63.53 grams. The cube of metal is placed into 64.7 mL of water. Determine the new volume of water if the density of the metal is 9.15 g / cm^3. Record your answer to three decimal places. Do not put units in your answer!

Solution Steps

To find the volume of the metal cube, we use the formula for density:

\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]

Rearranging the formula to solve for volume, we have:

\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} \]

Substituting the given values:

\[ \text{Volume} = \frac{63.53 \, \text{g}}{9.15 \, \text{g/cm}^3} = 6.9443 \, \text{cm}^3 \]

Since \(1 \, \text{cm}^3 = 1 \, \text{mL}\), the volume of the metal cube in milliliters is:

\[ 6.9443 \, \text{cm}^3 = 6.9443 \, \text{mL} \]

The new volume of water is the initial volume of water plus the volume of the metal cube:

\[ \text{New Volume} = 64.7 \, \text{mL} + 6.9443 \, \text{mL} = 71.6443 \, \text{mL} \]

Rounding to three decimal places, the new volume of water is:

\[ 71.644 \, \text{mL} \]

Final Answer