Question

when a 5.00 g piece of metal a was immersed in 38.0 ml of water, the water level rose to 50.0 ml. similarly, when a 5.00 g piece of metal b was immersed in 38.0 ml of water, the level of water rose to 60.0 ml. compare the density of metal a and metal b.

Explanation:

Step1: Calculate volume of Metal A

The volume of Metal A is the change in water level. Initial water volume is 38.0 mL, final is 50.0 mL. So volume \( V_A = 50.0 - 38.0 = 12.0 \, \text{mL} \)

Step2: Calculate density of Metal A

Density formula is \(
ho = \frac{m}{V} \). Mass of A is 5.00 g, volume is 12.0 mL. So \(
ho_A = \frac{5.00}{12.0} \approx 0.417 \, \text{g/mL} \)

Step3: Calculate volume of Metal B

Initial water volume 38.0 mL, final 60.0 mL. Volume \( V_B = 60.0 - 38.0 = 22.0 \, \text{mL} \)

Step4: Calculate density of Metal B

Mass of B is 5.00 g, volume 22.0 mL. \(
ho_B = \frac{5.00}{22.0} \approx 0.227 \, \text{g/mL} \)

Step5: Compare densities

Compare \(
ho_A \) and \(
ho_B \). \( 0.417 > 0.227 \), so \(
ho_A >
ho_B \)

Answer:

The density of Metal A (\( \approx 0.417 \, \text{g/mL} \)) is greater than the density of Metal B (\( \approx 0.227 \, \text{g/mL} \)).