Question

When 50 mL of water is added, the graduated cylinder has a mass of 180 g. If a rook is added to the graduated cylinder, the water level rises to 90 mL and the total mass is now 270 g. What is the density of the rock? ​

Answer 1

• ∆V=90-50=40mL
• ∆m=270-180=90g

Density be
`\rho`

`\\ \sf\longmapsto \rho=(m)/(v)`

`\\ \sf\longmapsto \rho=(90)/(40)`

`\\ \sf\longmapsto \rho=(9)/(4)`

`\\ \sf\longmapsto \rho=2.2g/ml`

Answer 2

2.25 (or
`(9)/(4)`) g/mL

Skills needed: Density

Explanation:

1) Briefly, let's cover what density is:

- It is
`(mass)/(volume)`

- Measures the mass per unit of volume

- In this situation the formula for density is:
`(m_f-m_i)/(v_f-v_i)`

--->
`m_f` is the final mass

--->
`m_i` is the initial mass

--->
`v_f` is the final volume

--->
`v_i` is the initial volume

2) In this case, we start out with a mass of 180 grams as stated in the problem, and also 50 mL of water initially.

This means that:

-
`m_i` is 180 g

-
`v_i` is 50 mL

3) Also, we end up with 270 g in mass, and 90 mL of water finally.

-
`m_f` is 270g

-
`v_f` is 90 mL

4) We can find density by substitute given the values above:

`(270-180)/(90-50) = (90)/(40) = (9)/(4) \text{ or } 2.25`

5) Our answer is 2.25 g/mL (since mass is grams (g), and volume is milliliters (mL) in the problem above).