In this question, we have to find the **mass** of the **fuel** with which the airplane took off from the runway while the **amount of fuel** in **liters **and the **density** are known. The basic concept behind this question is the knowledge of **density** and** mass**. We should know the difference between these two **physical quantities, **the formula for calculating **mass **and** density,** and the relationship between them as well.

In physics, **density** is represented as **mass per unit volume**. **Density** is represented by the symbol $\rho $, whereas in **mathematics** we can write it as **mass** being **divided** by the **volume**.

\[ Density = \dfrac{Mass}{Volume} \]

Which can also be written as:

\[ \displaystyle \rho = \dfrac{m}{V} \]

Here in this formula, we have:

$m\ =\ Mass \space of \space the \space object $

$V\ =\ Volume \space of \space the \space object $

$\rho\ =\ Density$

The **unit of density** will be the **unit of mass** **over the unit of volume,** which is defined as** grams per centimeters cube** $\dfrac {g}{cm^3 }$ or** kilograms per liter** $\dfrac {Kg}{L }$

In physics, the term **mass** implies how much **matter** is enclosed within an object.

**Mass** determines how much **inertia** is within the object, whereas** density** determines the** degree of compactness** (how close the atoms are together within the substance).

\[ Mass = Density \space \times \space Volume \]

Which can also be written as:

\[ m = \rho \space \times \space V \]

Here in this formula, we have:

$m\ =\ Mass \space of \space the \space object $

$V\ =\ Volume \space of \space the \space object $

$\rho\ =\ Density$

The unit of mass is **kilograms** $Kg $ or ** grams** $g $

## Expert Answer

Given in the question statement:

$Volume\ =\ V =\ 254 L =254 \times 10^3 mL$

$Density = \rho = 0.821$ $\dfrac { g}{ mL }$

Now to calculate the** mass,** we will use the formula:

\[m = \rho \space \times \space V \]

Now putting values in the above equation, we get:

\[m = 0.821 \times \space 245 \times 10^3\]

\[m=201145\ g\]

## Numerical Results

A small airplane took off with the **mass of fuel** to be $m= 201145g$ when the volume of the fuel was $254 L$ and the **density of the fuel** was $0.821$ $\dfrac { g}{ mL }$.

\[m = 201145\ g \]

## Example

A small plane takes on the fuel of $245 L$. If the **mass** is $201145 g$, calculate the** density** of the **fuel** in **grams per milliliter** with which the airplane has taken off.

Given in the question statement:

$Volume\ =\ V =\ 254 L=254 \times 10^3 mL$

$mass =\ m = 201145 g$

Now to calculate the **density,** we will use the formula:

\[\displaystyle \rho = \dfrac{m}{V} \]

Now putting values in the above equation, we get:

\[\rho =\dfrac{201145}{ 245 \times 10^3}\]

\[ Density = \rho = 0.821 \dfrac { g}{ mL }\]

Thus, the required** density** is:

\[\rho = 0.821 \dfrac { g}{ mL }\]