Question

1. a block of aluminum occupies a volume of 15.0 ml and weighs 40.5 g. what is its density?
2. mercury metal is poured into a graduated cylinder that holds exactly 22.5 ml. the mercury used to fill the cylinder weighs 306.0 g. from this information, calculate the density of mercury.
3. what is the weight of the ethyl alcohol that exactly fills a 200.0 ml container? the density of ethyl alcohol is 0.789 g/ml.
4. a rectangular block of copper metal weighs 1896 g. the dimensions of the block are 8.4 cm by 5.5 cm by 4.6 cm. from this data, what is the density of copper?
5. a flask that weighs 345.8 g is filled with 225 ml of carbon tetrachloride. the weight of the flask and carbon tetrachloride is found to be 703.55 g. from this information, calculate the density of carbon tetrachloride.
6. calculate the density of sulfuric acid if 35.4 ml of the acid weighs 65.14 g
7. find the mass of 250.0 ml of benzene. the density of benzene is 0.8765 g/ml.
8. a block of lead has dimensions of 4.50 cm by 5.20 cm by 6.00 cm. the block weighs 1587 g. from this information, calculate the density of lead.
9. 28.5 g of iron shot is added to a graduated cylinder containing 45.50 ml of water. the water level rises to the 49.10 ml mark, from this information, calculate the density of iron.
10. what volume of silver metal will weigh exactly 2500.0 g. the density of silver is 10.5 g/cm³.

Explanation:

Let's solve each problem one by one using the density formula \( d = \frac{m}{v} \), \( m = d \times v \), or \( v = \frac{m}{d} \) as needed.

Problem 1:

Step 1: Identify formula and values
Density formula: \( d = \frac{m}{v} \)
\( m = 40.5 \, \text{g} \), \( v = 15.0 \, \text{mL} \)

Step 2: Calculate density
\( d = \frac{40.5}{15.0} = 2.7 \, \text{g/mL} \)

Problem 2:

Step 1: Identify formula and values
\( d = \frac{m}{v} \), \( m = 306.0 \, \text{g} \), \( v = 22.5 \, \text{mL} \)

Step 2: Calculate density
\( d = \frac{306.0}{22.5} = 13.6 \, \text{g/mL} \)

Problem 3:

Step 1: Identify formula and values
Mass formula: \( m = d \times v \)
\( d = 0.789 \, \text{g/mL} \), \( v = 200.0 \, \text{mL} \)

Step 2: Calculate mass
\( m = 0.789 \times 200.0 = 157.8 \, \text{g} \)

Problem 4:

Step 1: Calculate volume of the block
Volume of a rectangular block: \( v = l \times w \times h \)
\( l = 8.4 \, \text{cm} \), \( w = 5.5 \, \text{cm} \), \( h = 4.6 \, \text{cm} \)
\( v = 8.4 \times 5.5 \times 4.6 = 212.52 \, \text{cm}^3 = 212.52 \, \text{mL} \) (since \( 1 \, \text{cm}^3 = 1 \, \text{mL} \))

Step 2: Calculate density
\( d = \frac{m}{v} \), \( m = 1896 \, \text{g} \), \( v = 212.52 \, \text{mL} \)
\( d = \frac{1896}{212.52} \approx 8.92 \, \text{g/mL} \)

Problem 5:

Step 1: Find mass of carbon tetrachloride
Mass of flask + CCl₄ = 703.55 g, Mass of flask = 345.8 g
Mass of CCl₄: \( m = 703.55 - 345.8 = 357.75 \, \text{g} \)

Step 2: Calculate density
\( d = \frac{m}{v} \), \( v = 225 \, \text{mL} \), \( m = 357.75 \, \text{g} \)
\( d = \frac{357.75}{225} = 1.59 \, \text{g/mL} \)

Problem 6:

Step 1: Identify formula and values
\( d = \frac{m}{v} \), \( m = 65.14 \, \text{g} \), \( v = 35.4 \, \text{mL} \)

Step 2: Calculate density
\( d = \frac{65.14}{35.4} \approx 1.84 \, \text{g/mL} \)

Problem 7:

Step 1: Identify formula and values
\( m = d \times v \), \( d = 0.8765 \, \text{g/mL} \), \( v = 250.0 \, \text{mL} \)

Step 2: Calculate mass
\( m = 0.8765 \times 250.0 = 219.125 \, \text{g} \)

Problem 8:

Step 1: Calculate volume of the block
\( v = l \times w \times h = 4.50 \times 5.20 \times 6.00 = 140.4 \, \text{cm}^3 = 140.4 \, \text{mL} \)

Step 2: Calculate density
\( d = \frac{m}{v} \), \( m = 1587 \, \text{g} \), \( v = 140.4 \, \text{mL} \)
\( d = \frac{1587}{140.4} \approx 11.3 \, \text{g/mL} \)

Problem 9:

Step 1: Find volume of iron shot
Volume of water + iron = 49.10 mL, Volume of water = 45.50 mL
Volume of iron: \( v = 49.10 - 45.50 = 3.60 \, \text{mL} \)

Step 2: Calculate density
\( d = \frac{m}{v} \), \( m = 28.5 \, \text{g} \), \( v = 3.60 \, \text{mL} \)
\( d = \frac{28.5}{3.60} \approx 7.92 \, \text{g/mL} \)

Problem 10:

Step 1: Identify formula and values
Volume formula: \( v = \frac{m}{d} \)
\( m = 2500.0 \, \text{g} \), \( d = 10.5 \, \text{g/cm}^3 \) (since \( 1 \, \text{cm}^3 = 1 \, \text{mL} \))

Step 2: Calculate volume
\( v = \frac{2500.0}{10.5} \approx 238.1 \, \text{cm}^3 \)

Answer:

s:

1. \( \boldsymbol{2.7 \, \text{g/mL}} \)
2. \( \boldsymbol{13.6 \, \text{g/mL}} \)
3. \( \boldsymbol{157.8 \, \text{g}} \)
4. \( \boldsymbol{\approx 8.92 \, \text{g/mL}} \)
5. \( \boldsymbol{1.59 \, \text{g/mL}} \)
6. \( \boldsymbol{\approx 1.84 \, \text{g/mL}} \)
7. \( \boldsymbol{219.125 \, \text{g}} \)
8. \( \boldsymbol{\approx 11.3 \, \text{g/mL}} \)
9. \( \boldsymbol{\approx 7.92 \, \text{g/mL}} \)
10. \( \boldsymbol{\approx 238.1 \, \text{cm}^3} \)