Question

A drop of acetone (nail polish remover) has a mass of 35 mg and a density of 0.788 g/cm3.what is its volume in cubic centimeters?

Answer 1

The volume of acetone is 0.044 cm3

$\boxed{\boxed{\bold{Further~explanation}}}$

Density is a quantity derived from the mass and volume

density is the ratio of mass per unit volume

With the same mass, the volume of objects that have a high density will be smaller than type

The unit of density can be expressed in g / cm3 or kg / m3

Density formula:

$\large{\boxed{\rho~=~\frac{m}{V} }}}$

ρ = density

m = mass

v = volume

A common example is the water density of 1 gr / cm3

The mass itself is often equated with weight, even though it is different. Mass is the amount of matter in the matter while weight is related to the gravitational force

Known variable

mass of 35 mg

a density of 0.788 g / cm3

Asked

volume in cubic centimeters

Answer

mass is known to be 35 mg, then we change it first to units of gram (g) adjusting to units of density

35 mg = 35.10-3 grams

then:

volume = mass / density

volume = 35.10-3 / 0.788

volume = 0.044 cm3

$\boxed{\boxed{\bold{Learn~more}}}$

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Keywords: density, mass, volume, A drop of acetone

Answer 2

The volume of the drop of acetone is $\boxed{0.0444\,{\text{c}}{{\text{m}}^{\text{3}}}}$.

Further Explanation:

Given:

The density of the drop of acetone is $0.788\,{{\text{g}} \mathord{\left/{\vphantom {{\text{g}} {{\text{c}}{{\text{m}}^{\text{3}}}}}} \right.\kern-\nulldelimiterspace} {{\text{c}}{{\text{m}}^{\text{3}}}}}$.

The mass of the drop of acetone is $35\,{\text{mg}}$.

Concept:

The density of the drop is the amount of matter contained in the unit volume of the drop. The density is defined as the mass per unit volume. The unit of density of a particular object is represent as ${{\text{g}} \mathord{\left/ {\vphantom {{\text{g}} {{\text{c}}{{\text{m}}^{\text{3}}}}}} \right.\kern-\nulldelimiterspace} {{\text{c}}{{\text{m}}^{\text{3}}}}}$ or ${{{\text{kg}}} \mathord{\left/{\vphantom {{{\text{kg}}} {{{\text{m}}^{\text{3}}}}}} \right.\kern-\nulldelimiterspace} {{{\text{m}}^{\text{3}}}}}$.

The mathematical representation of the density is as shown below:

$\boxed{\rho = \frac{m}{V}}$

Here, $\rho$ is the density of the acetone, $m$ is the mass of the drop and $V$ is the volume contained.

Convert the mass of the drop of acetone into gram.

$\begin{aligned}m&= 35\,{\text{mg}}\left( {\frac{{1\,{\text{g}}}}{{1000\,{\text{mg}}}}} \right)\\&= 0.035\,{\text{g}}\\\end{aligned}$  

Substitute $0.788\,{{\text{g}} \mathord{\left/ {\vphantom {{\text{g}} {{\text{c}}{{\text{m}}^{\text{3}}}}}} \right. \kern-\nulldelimiterspace} {{\text{c}}{{\text{m}}^{\text{3}}}}}$ for $\rho$ and $0.035\,{\text{g}}$ for $V$ in above expression.

$\begin{aligned}0.788\,{{\text{g}} \mathord{\left/{\vphantom {{\text{g}} {{\text{c}}{{\text{m}}^{\text{3}}}}}} \right.\kern-\nulldelimiterspace} {{\text{c}}{{\text{m}}^{\text{3}}}}} &= \frac{{0.035\,{\text{g}}}}{V} \\V&= \frac{{0.035\,{\text{g}}}}{{0.788\,{{\text{g}} \mathord{\left/{\vphantom {{\text{g}} {{\text{c}}{{\text{m}}^{\text{3}}}}}} \right.\kern-\nulldelimiterspace} {{\text{c}}{{\text{m}}^{\text{3}}}}}}}\\&= 0.0444\,{\text{c}}{{\text{m}}^{\text{3}}} \\\end{aligned}$  

Thus, the volume of the drop of acetone is $\boxed{0.0444\,{\text{c}}{{\text{m}}^{\text{3}}}}$.

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Answer Details:

Grade: High School

Chapter: Units and measurement

Subject: Physics

Keywords: Drop, acetone, nail polish remover, 35 mg, density, mass, volume, cubic centimetres, mass per unit volume.