Question

A graduated cylinder contains 155 ml of water. a 15.0-g piece of iron (density = 7.86 g/ml) and a 20.0-g piece of lead (density = 11.3 g/ml) are added. what is the new water level, in milliliters, in the cylinder?

Answer 1

Initial volume of water is 155 mL. Convert the mass of iron and lead into volume using density values as follows:

d=\frac{m}{V}

Here, d is density, m is mass and V is volume of objects

volume of  piece of iron can be calculated as:

V=\frac{m}{d}=\frac{15 g}{7.86 g/mL}=1.90 mL

Similarly, volume of lead can be calculated as:

V=\frac{m}{d}=\frac{20 g}{11.3g/mL}=1.77 mL

To calculate the final volume of water, volume of both iron and lead should be added to the initial volume of water.

V_{final}=V_{initial}+V_{iron}+V_{lead}=155 mL+1.90 mL+1.77mL=158.67 mL

Thus, new water level in milliliters will be 158.67 mL.