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 $\rho$

$\\ \sf\longmapsto \rho=\dfrac{m}{v}$

$\\ \sf\longmapsto \rho=\dfrac{90}{40}$

$\\ \sf\longmapsto \rho=\dfrac{9}{4}$

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

Answer 2

Answer:

2.25 (or $\frac{9}{4}$) g/mL

Skills needed: Density

Step-by-step explanation:

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

- It is $\frac{mass}{volume}$

- Measures the mass per unit of volume

- In this situation the formula for density is: $\frac{m_f-m_i}{v_f-v_i}$

---> $m_f$ 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:

$\frac{270-180}{90-50} = \frac{90}{40} = \frac{9}{4} \text{ or } 2.25$

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