Let's solve each problem step by step using the formula for density:
$$
\text{Density} = \frac{\text{Mass}}{\text{Volume}} \quad \text{or} \quad d = \frac{M}{V}
$$
And its rearranged forms:
- $ M = d \times V $
- $ V = \frac{M}{d} $
---
1. Find the unknown quantity
#### a) $ d = 3 \, \text{g/mL}, \, V = 100 \, \text{mL}, \, M = ? $
Use:
$$
M = d \times V = 3 \, \text{g/mL} \times 100 \, \text{mL} = 300 \, \text{g}
$$
✔ Answer: 300 g
---
#### b) $ d = ?, \, V = 95 \, \text{mL}, \, M = 950 \, \text{g} $
Use:
$$
d = \frac{M}{V} = \frac{950 \, \text{g}}{95 \, \text{mL}} = 10 \, \text{g/mL}
$$
✔ Answer: 10 g/mL
---
#### c) $ d = 5 \, \text{g/cm}^3, \, M = 20 \, \text{g}, \, V = ? $
Use:
$$
V = \frac{M}{d} = \frac{20 \, \text{g}}{5 \, \text{g/cm}^3} = 4 \, \text{cm}^3
$$
✔ Answer: 4 cm³
---
2. Find the unknown quantity
#### a) $ d = 24 \, \text{g/mL}, \, V = 2 \, \text{mL}, \, M = ? $
$$
M = d \times V = 24 \times 2 = 48 \, \text{g}
$$
✔ Answer: 48 g
---
#### b) $ d = ?, \, V = 10 \, \text{mL}, \, M = 150 \, \text{g} $
$$
d = \frac{150}{10} = 15 \, \text{g/mL}
$$
✔ Answer: 15 g/mL
---
#### c) $ d = ?, \, V = 50 \, \text{mL}, \, M = 500 \, \text{g} $
$$
d = \frac{500}{50} = 10 \, \text{g/mL}
$$
✔ Answer: 10 g/mL
---
##
WORD PROBLEMS
1. A block of aluminum occupies a volume of 15 mL and weighs 45 g. What is its density?
$$
d = \frac{M}{V} = \frac{45 \, \text{g}}{15 \, \text{mL}} = 3 \, \text{g/mL}
$$
✔ Answer: 3 g/mL
---
2. Mercury metal is poured into a graduated cylinder that holds exactly 20 mL. The mercury used to fill the cylinder weighs 400 g. Calculate the density of mercury.
$$
d = \frac{400 \, \text{g}}{20 \, \text{mL}} = 20 \, \text{g/mL}
$$
✔ Answer: 20 g/mL
---
3. A rectangular block of metal weighs 3200 g. Dimensions are 8 cm × 5 cm × 4 cm. Find density.
First, find volume:
$$
V = \text{length} \times \text{width} \times \text{height} = 8 \times 5 \times 4 = 160 \, \text{cm}^3
$$
Now calculate density:
$$
d = \frac{3200 \, \text{g}}{160 \, \text{cm}^3} = 20 \, \text{g/cm}^3
$$
✔ Answer: 20 g/cm³
---
4. What volume of silver metal will weigh exactly 2500.0 g? Density of silver = 10.5 g/cm³.
Use:
$$
V = \frac{M}{d} = \frac{2500.0 \, \text{g}}{10.5 \, \text{g/cm}^3} \approx 238.095 \, \text{cm}^3
$$
Rounded to one decimal place:
238.1 cm³
✔ Answer: 238.1 cm³
---
5. A block of lead has dimensions 4.5 cm × 5.2 cm × 6.0 cm. Weight = 1587 g. Find density.
Step 1: Volume
$$
V = 4.5 \times 5.2 \times 6.0 = (4.5 \times 5.2) = 23.4; \quad 23.4 \times 6.0 = 140.4 \, \text{cm}^3
$$
Step 2: Density
$$
d = \frac{1587 \, \text{g}}{140.4 \, \text{cm}^3} \approx 11.30 \, \text{g/cm}^3
$$
✔ Answer: 11.3 g/cm³
---
##
✔ Final Answers Summary
Part 1:
a) M =
300 g
b) d =
10 g/mL
c) V =
4 cm³
Part 2:
a) M =
48 g
b) d =
15 g/mL
c) d =
10 g/mL
Word Problems:
1.
3 g/mL
2.
20 g/mL
3.
20 g/cm³
4.
238.1 cm³
5.
11.3 g/cm³
Let me know if you'd like this formatted as a printable answer key!
Parent Tip: Review the logic above to help your child master the concept of density calculations worksheet.
