# Challenging Density Problems?

Please answer as many of the questions as you can, showing how you got to your answer. Thanks.

1. At a cost of $350/oz., how much would you have to pay for 1.00 cubic foot of solid gold? (1 oz = 28.4 g; 1 in = 2.54 cm)

2. A hollow lead pipe is 25.0 cm long and has an OD (outer diameter) of 3.25 cm. It weighs 562 g. What is its ID (inner diameter)?

3. A 25.0 L helium balloon floats up to the ceiling. At 12:00 noon, it springs a slow leak that lets 1.00 L of helium leak out each hour. At what time will it begin to descend to the floor? (The rubber of the balloon weighs 8.20 g).

4. A coin is made of gold plated on copper. The coin's density is precisely 10.0 g/ml. What is the percent by mass of copper in the coin?

Any help is appreciated.

### 1 Answer

- electron1Lv 76 years agoFavorite Answer
1. At a cost of $350/oz., how much would you have to pay for 1.00 cubic foot of solid gold? (1 oz = 28.4 g; 1 in = 2.54 cm)

Let’s determine cost of 1 gram of gold.

Cost = $(350 ÷ 28.4)/g

The cost per gram is approximately $12.32

Let’s determine the mass of gold.

Density of gold = 19.3 g/cm^3

Let’s convert 1 ft^3 to cm^3

1 ft = 12 in, 12 in = 12 * 2.54 = 30.48 cm

1 ft^3 = 30.48^3 cm^3

Mass of gold = 19.3 * 30.48^3 grams

The mass of gold is approximately 5.164 * 10^4 g

Cost = (350 ÷ 28.4) * (19.3 * 30.48^3)

The cost is approximately $6.735 * 10^5

2. A hollow lead pipe is 25.0 cm long and has an OD (outer diameter) of 3.25 cm. It weighs 562 g. What is its ID (inner diameter)?

To determine the inner diameter we need to know the cross sectional area of the inside of the pipe.

Cross sectional area of the inside of the pipe = Cross sectional area of the pipe – Cross sectional area of the lead part.

Cross sectional area of pipe = π * d = π * 3.25

Cross sectional area = Volume ÷ Length

Let’s determine the volume of lead.

Density of lead = 11.4 g/cm^3

Volume of lead = 562 ÷ 11.4 ≈ 49.3 cm^3

Cross sectional area of lead = 562 ÷ 11.4 ÷ 25 = 22.48 ÷ 11.4 ≈ 1.972 cm^2

Cross sectional area of inside = π * 3.25 – (22.48 ÷ 11.4) ≈ 8.2 cm^2

To determine the inner diameter, divide the area by π

d = (π * 3.25 – 22.48 ÷ 11.4) ÷ π

The diameter is approximately 2.62 cm.

3. A 25.0 L helium balloon floats up to the ceiling. At 12:00 noon, it springs a slow leak that lets 1.00 L of helium leak out each hour. At what time will it begin to descend to the floor? (The rubber of the balloon weighs 8.20 g).

As the volume of the balloon decreases, the buoyant force decreases. When the buoyant force is less than the total weight of the balloon, the balloon will begin to descend.

Buoyant force = Density * g * V

Density of air at STP = 1.29 g/ml = 1.29 kg/L

Buoyant force = 1.29 * 9.8 * 25 = 316.05 N

Weight of rubber = 0.082 * 9.8 = 0.8036 N

Let’s determine the weight of 25 liters of He.

Density of He = 0.1785 g/ml = 0.1785 kg/L

Mass = 0.1785 * 25 * 9.8 = 43.7325 N

Total weight of balloon = 44.5361

Let’s determine the weight of 1 liter of He.

Weight = 0.1785 * 1 * 9.8 = 1.7493 N

This is the mass He that is leaking out of the balloon each hour. The buoyant force will decrease this much each hour, until the buoyant force is equal to the total weight of the balloon.

316.05 – 1.7493 * n = 44.5361

Subtract 316.05 from both sides.

-1.7493 * n = -271.5139

Divide both sides by 1.7493

The number of hours is approximately 155.2

4. A coin is made of gold plated on copper. The coin's density is precisely 10.0 g/ml. What is the percent by mass of copper in the coin?

Let’s assume that the mass of the coin = 5 g

Volume of coin = 5 g/ 10 g/ml = 0.5 ml

The mass of the coin is equal to the sum of the mass of copper and gold.

Mass of copper = 8.96 * Vcu

Mass of gold = 19.3 * Vau

Mass of copper + gold = Mass of coin

Eq#1: 8.96 * Vcu + 19.3 * Vau = 5

The volume of the coin is equal to the sum of the volume of copper and gold.

Eq#2: Vcu + Vau = 0.5

Solve these two equations for Vcu and V au. Then determine the mass of Cu and Au.

Vau = 0.5 – Vcu

Substitute 0.5 – Vcu for Vau in Eq#1.

8.96 * Vcu + 19.3 * (0.5 – Vcu) = 5

8.96 * Vcu + 9.65 – 19.3 * Vcu = 5

-10.34 * Vcu = -4.65

Vcu = 4.65 ÷ 10.34 ≈ 0.4497 ml

Mass of Cu = 8.96 * (4.65 ÷ 10.34) ≈ 4.0294 g

Vau = 0.5 – Vcu = 0.5 – 4.65 ÷ 10.34 ≈ 0.0503 ml

Mass of Au = 19.3 * (0.5 – 4.65 ÷ 10.34) ≈ 0.9706 kg

% Cu = 100 * 4.0294/5

% Au = 100 * 0.9706 /5

Thank you kindly darling