Question

a graduated cylinder contains 20.5ml of water. what is the new water level after 41.3g of silver metal with a density of 10.5g/cm3 is submerged in the water

Answer 1

The increase in water level must equal the volume of the submerged silver.

 

ΔV = volume of silver

 

$ΔV = \frac{m}{ρ }=\frac {(41.6 g)}{(10.5 g/mL) }=3.96 mL$

V = final water level

 

$V = (25.0 + 3.96) mL=28.96 mL$