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What is the density (g/mL) of each of the following samples? a. An cbony carving has a mass of \(275 \mathrm{~g}\) and a volume of \(207 \mathrm{~cm}^{3}\) b. A \(14.3-\mathrm{cm}^{3}\) sample of tin has a mass of \(0.104 \mathrm{~kg}\). c. A bottle of acetone (fingernail polish remover) contains \(55.0 \mathrm{~mL}\) of acetone with a mass of \(43.5 \mathrm{~g}\).

Short Answer

Expert verified
a) 1.33 g/cm3, b) 7.27 g/cm3, c) 0.791 g/mL

Step by step solution

01

Understand the Formula for Density

Density (\rho) is calculated using the formula: \[ \rho = \frac{\text{Mass}}{\text{Volume}} \] where the mass is in grams (g) and the volume is in milliliters (mL) or cubic centimeters (\text{cm}^3). Remember that 1 mL = 1 cm^3.
02

Calculate Density for Sample (a)

For the ebony carving: \[ \text{Mass} = 275 \, \text{g}, \text{Volume} = 207 \, \text{cm}^3 \] Use the formula \[ \rho = \frac{275 \, \text{g}}{207 \, \text{cm}^3} = 1.33 \, \text{g/cm}^3 \]
03

Convert Units for Sample (b) if Needed

For the tin sample: \[ \text{Mass} = 0.104 \, \text{kg} = 104 \, \text{g}, \text{Volume} = 14.3 \, \text{cm}^3 \] Convert the mass from kilograms to grams: \[ 0.104 \, \text{kg} \times 1000 = 104 \, \text{g} \]
04

Calculate Density for Sample (b)

Use the formula for density: \[ \rho = \frac{104 \, \text{g}}{14.3 \, \text{cm}^3} \approx 7.27 \, \text{g/cm}^3 \]
05

Calculate Density for Sample (c)

For the acetone: \[ \text{Mass} = 43.5 \, \text{g}, \text{Volume} = 55.0 \, \text{mL} \] Use the formula: \[ \rho = \frac{43.5 \, \text{g}}{55.0 \, \text{mL}} = 0.791 \, \text{g/mL} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

mass and volume relationship
Density is a unique property of matter that relates its mass and volume. Mass refers to the amount of matter in an object and is often measured in grams (g) or kilograms (kg). Volume refers to how much space the object occupies and is usually measured in cubic centimeters (\text{cm}^3) or milliliters (mL). The relationship between mass and volume is described by the formula for density, given by \( \rho = \frac{\text{Mass}}{\text{Volume}} \). This means that density (\textGreek{\rho}) tells you how much mass is in a given volume of a substance.

An easy way to understand this is to imagine two objects: one made of a heavy metal like iron and one made of wood. Even if they both have the same volume, the iron will have more mass and hence a higher density because the particles are packed more closely together.

  • Example: In sample (a), an ebony carving has a mass of 275 g and a volume of 207 cm³.
  • This makes the density calculation simple:
  • \( \rho = \frac{275 \, \text{g}}{207 \, \text{cm}^3} = 1.33 \, \text{g/cm}^3 \)
unit conversion
Converting units is essential in calculations because it ensures that all measurements are in the same units. When dealing with density, you often need to convert between kilograms and grams, or milliliters and cubic centimeters (\text{cm}^3).

Here are some common conversions to remember:
  • 1 kilogram (kg) = 1000 grams (g)
  • 1 milliliter (mL) = 1 cubic centimeter (\text{cm}^3)
For example, in sample (b), we have a tin sample with a mass given in kilograms, which needs to be converted to grams:
  • Mass = 0.104 kg = 104 g
After converting, we use this mass to compute the density:
  • \( \rho = \frac{104 \, \text{g}}{14.3 \, \text{cm}^3} = 7.27 \, \text{g/cm}^3 \)
Converting units makes it easier to use the density formula consistently and avoid errors.
density formula
The density formula \( \rho = \frac{\text{Mass}}{\text{Volume}} \) is used to determine how much mass is contained in a given volume. This formula is vital when comparing materials or substances. The SI unit for density is kilograms per cubic meter (kg/m³), but for smaller substances or laboratory settings, we often use grams per milliliter (g/mL) or grams per cubic centimeter (g/cm³).

For example, in sample (c), the mass of acetone is given as 43.5 g, and the volume is 55.0 mL. Using the density formula, we get:
  • \( \rho = \frac{43.5 \, \text{g}}{55.0 \, \text{mL}} = 0.791 \, \text{g/mL} \)
By using this simple calculation, we can compare the density of acetone to other substances and understand its properties better. Knowing the density helps in numerous practical applications, from determining material types to quality control in manufacturing.

Remember to always check that your mass and volume are in compatible units before applying the formula.

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Most popular questions from this chapter

Perform each of the following conversions using metric conversion factors: a. \(44.2 \mathrm{~mL}\) to liters b. \(8.65 \mathrm{~m}\) to nanometers c. \(5.2 \times 10^{\mathrm{8}} \mathrm{g}\) to megagrams d. \(0.72 \mathrm{ks}\) to milliseconds

In which of the following pairs do both numbers contain the same number of significant figures? a. \(11.0 \mathrm{~m}\) and \(11.00 \mathrm{~m}\) b. \(0.0250 \mathrm{~m}\) and \(0.205 \mathrm{~m}\) c. \(0.00012 \mathrm{~s}\) and \(12000 \mathrm{~s}\) d. \(250.0 \mathrm{~L}\) and \(2.5 \times 10^{-2} \mathrm{~L}\)

For each of the following pairs, which is the larger unit? a. milligram or kilogram b. milliliter or microliter c. \(\mathrm{m}\) or \(\mathrm{km}\) d. \(\mathrm{kL}\) or \(\mathrm{dL}\) e. nanometer or picometer

Perform each of the following calculations, and give an answer with the correct number of decimal places: a. \(5.08 \mathrm{~g}+25.1 \mathrm{~g}\) b. \(85.66 \mathrm{~cm}+104.10 \mathrm{~cm}+0.025 \mathrm{~cm}\) c. \(24.568 \mathrm{~mL}-14.25 \mathrm{~mL}\) d. \(0.2654 \mathrm{~L}-0.2585 \mathrm{~L}\)

Write the equality and two conversion factors, and identify the numbers as exact or give the number of significant figures for each of the following: a. The label on a bottle reads \(10 \mathrm{mg}\) of furosemide per \(1 \mathrm{~mL}\). b. The Daily Value (DV) for selenium is \(70 . \mathrm{mcg}\). c. An IV of normal saline solution has a flow rate of \(85 \mathrm{~mL}\) per hour. d. One capsule of fish oil contains \(360 \mathrm{mg}\) of omega-3 fatty acids.

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