Question

title = q8a8 you have a piece of gold with a mass of 483.0 grams. you have a graduated cylinder containing 50.0 ml of water. you add the piece of gold to the graduated cylinder, it sinks to the bottom, and the water level rises to read 75.0 ml. what is the density of the gold?

○ 19.3 g/cm³

○ 6.44 g/cm³

○ 1.45 × 10² g/cm³

○ 161 g/cm³

○ 7.78 × 10⁴ g/cm³

Explanation:

Step1: Find the volume of gold

The volume of the gold is equal to the change in water volume in the graduated cylinder. Initial water volume is \( 50.0 \, \text{mL} \), final volume is \( 75.0 \, \text{mL} \). So volume \( V = 75.0 - 50.0 = 25.0 \, \text{mL} \). Since \( 1 \, \text{mL} = 1 \, \text{cm}^3 \), \( V = 25.0 \, \text{cm}^3 \).

Step2: Calculate density

Density formula is \(
ho = \frac{m}{V} \), where \( m = 483.0 \, \text{g} \), \( V = 25.0 \, \text{cm}^3 \). So \(
ho = \frac{483.0}{25.0} = 19.32 \approx 19.3 \, \text{g/cm}^3 \).

Answer:

A. \( 19.3 \, \text{g/cm}^3 \)