Question

A graduated cylinder contains 23.0mL {\rm mL} of water. What is the new water level after 36.2g {\rm g} of silver metal with a density of 10.5 g/mL {\rm g/mL} is submerged in the water?

Answer 1

A graduated cylinder contains 23.0 mL $\{\mathrm{rm~mL}\}$ of water. What is the new water level after 36.2 g $\{\mathrm{rm~g}\}$ of silver metal with a density of 10.5 g/mL $\{\mathrm{rm~g/mL}\}$ is submerged in the water?

Solution

d = m/V; V = m/d; where: d – density, m – mass, V – volume.

So, V = 36.2 / 10.5 = 3.45 mL of silver

A new water level will be 23 + 3.45 = 26.45 mL