Q. 2 - An Introduction to General, Organic, and Biological Chemistry

Question

The mileage for a motorcycle with a fuel-tank capacity of $22 \mathrm{L}$ is $35 \mathrm{mi}/\mathrm{gal}$.
a. How long a trip, in kilometers, can be made on one full tank of gasoline?
b. If the price of gasoline is $\$3.82$ per gallon, what would be the cost of fuel for the trip?
c. If the average speed during the trip is $44 \mathrm{mi}/\mathrm{h}$ how many ours will it take to reach the destination?
d. If the density of gasoline is $0.74 \mathrm{g}/\mathrm{mL}$, what is the mass, in grams, of the fuel in the tank?
e. When $1.00 \mathrm{g}$ of gasoline burns, $47 \mathrm{kJ}$ of energy is released. How many kilojoules are produced when the fuel in one full tank is burned?

Step-by-Step Solution

Answer

Part(a) $327.25 \mathrm{Km}$long trip can be made on full tank of gasoline.

Part(b) $\$18.5$ is cost of fuel for the trip.

Part(c) Time taken to complete the trip is $4.622 \mathrm{h}$

Part(d) $16280 \mathrm{g}$ is the mass of fuel in the tank.

Part(e) $1034000 \mathrm{kJ}$Energy is released when the fuel in one full tank is burned .

Part(a) Step 1: Given information

We are given mileage of motorcycle with fuel tank capacity.

Part(a) Step 2 : Simplify

As we know $1 \mathrm{mi}/\mathrm{gal} = 0.425 \mathrm{Km}/\mathrm{L}$

So, $35 \mathrm{mi}/\mathrm{gal} = 35\times 0.425 \mathrm{Km}/\mathrm{L}= 14.875 \mathrm{Km}/\mathrm{L}$

Fuel-tank capacity is $22 \mathrm{L}$

Therefore, $14.875 \mathrm{Km}/\mathrm{L}$ means in $22 \mathrm{L}$ motorcycle can cover a trip of $14.875 \frac{\mathrm{Km}}{\cancel{\mathrm{L}}}\times 22 \cancel{\mathrm{L}}= 327.25 \mathrm{Km}$

Part(b) Step 1: Given information

We are given mileage of motorcycle with fuel tank capacity.

Part(b) Step 2 : Simplify

As we know $1 \mathrm{gal} = 3.875 \mathrm{L}$

Therefore, $22 \mathrm{L} = 4.84 \mathrm{gal}$

The price of gasoline is $\$3.82 \mathrm{per} \mathrm{gal}$

So, $4.84 \mathrm{gal} = 4.84\cancel{ \mathrm{gal}} \times \frac{\$3.82}{\cancel{\mathrm{gal}}} = \$18.5$

Part(c) Step 1: Given information

We are given mileage of motorcycle with fuel tank capacity.

Part(c) Step 2 : Simplify

As we know, $1 \mathrm{mi}/\mathrm{h} = 1.609 \mathrm{Km}/\mathrm{h}$

therefore, $44 \mathrm{mi}/\mathrm{h} = 44 \times 1.609 \frac{\mathrm{Km}}{\mathrm{h}}=70.8 \mathrm{Km}/\mathrm{h}$

Average speed is $70.8 \mathrm{Km}/\mathrm{h}$

Total distance = $327.5 \mathrm{Km}$

so, $\frac{\mathrm{Distance}}{\mathrm{Time}}=\mathrm{average} \mathrm{speed}$

$$\begin{gathered} 70.8 \cancel{\mathrm{Km}}/\mathrm{h}=\frac{327.25 \cancel{\mathrm{Km}}}{\mathrm{time}} \\ \mathrm{time}=4.622 \mathrm{h} \end{gathered}$$

Time taken to complete the trip is $4.622 \mathrm{h}$

Part(d) Step 1 : Given information

We are given mileage of motorcycle with fuel tank capacity.

Part(d) Step 2 : Simplify

Density of gasoline is $0.74 \mathrm{g}/\mathrm{ml}$

Capacity of tank is $22 \mathrm{L} = 22000 \mathrm{mL}$

Density is mass divided by volume,

Therefore mass of fuel is $0.74 \frac{\mathrm{g}}{\cancel{\mathrm{mL}}}=\frac{\mathrm{mass}}{22000 \cancel{\mathrm{mL}}}= 16280 \mathrm{g}$

Part(e) Step 1: Given information

We are given mileage of motorcycle with fuel tank capacity.

Part(e) Step 2 : Simplify

As given, Energy produced is $47 \mathrm{kJ}/\mathrm{g}$

so, in full fuel tank energy produced will be $$\begin{gathered} 1 \mathrm{g} = 47 \mathrm{kJ} \\ 22000 \mathrm{g} = 1034000 \mathrm{kJ} \end{gathered}$$