A drop of gasoline has a mass of 22 mg and a density of 0.754 g/cm3. What is its volume in cubic centimeters?
Answer:
volume = 0.029 cm³
Explanation:
We are given that:
mass = 22 mg
We know that:
1 mg = 0.001 grams
This means that:
mass = 22 mg = 0.022 grams
We are also given that:
density = 0.754 g/cm³
Density is calculated using the following rule:
\(Density=\frac{mass}{volume}\)
Rewrite the equation to solve for volume:
\(volume = \frac{mass}{density}\)
Now, substitute with the values we have:
\(volume = \frac{0.022}{0.754} = 0.029 cm^3\)
Hope this helps :)
The volume of gasoline (in cubic centimeters) is \(\boxed{0.029\;\text{cm}^3}\).
Further Explanation:
Density is defined as the mass of the substance per unit volume of the substance. It is denoted by \(\rho\). It is generally expressed in the units of grams per cubic centimeter \((\text{g/cm}^3)\). The formula to calculate the density of a substance is as follows:
\(\rho=\dfrac{\text{M}}{\text{V}}\)
Here,
\(\rho\) is the density of the substance.
M is the mass of the substance.
V is the volume of the substance.
The formula to calculate the density of gasoline is as follows:
\(\text{Denstiy of gasoline}=\dfrac{\text{Mass of gasoline}}{\text{Volume of gasoline}}\) …… (1)
Rearrange equation (2) to calculate the volume of gasoline.
\(\text{volume of gasoline}=\dfrac{\text{Mass of gasoline}}{\text{Density of gasoline}}\) …… (2)
The mass of gasoline is 22 mg. This is to be converted into g. The conversion factor for this is,
\(1\;\text{mg}=10^{-3}\text{g}\)
Therefore the mass of gasoline can be calculated as follows:
\(\begin{aligned}{\text{Mass of gasoline}}&= \left( {22{\text{ mg}}} \right)\left( {\frac{{{{10}^{ - 3}}{\text{ g}}}}{{1{\text{ mg}}}}} \right)\\&= 0.022{\text{ g}} \\ \end{aligned}\)
The mass of gasoline is 0.022 g.
The density of gasoline is \(0.754\;\text{g/cm}^3\).
Substitute these values in equation (2).
\(\begin{aligned}\text{Volume of gasoline}&=\frac{0.022\;\text{g}}{0.754\;\text{cm}^3}\\&=0.029\text{cm}^3\\\end{aligned}\)
Therefore the volume of gasoline is \(0.029\;\text{cm}^3\).
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Answer details:
Grade: Middle School
Chapter: Density
Subject: Chemistry
Keywords: density, mass, volume, 22 mg, 0.022 g, 0.029 cm^3, volume of gasoline, mass of gasoline, density of gasoline, \(0.754 \;\text{g/cm}^3, g\), conversion factor.
