Answer:
27,397.23 L would be needed for a successful trip.
Explanation:
The problem gives us the density (ρ) of the fuel, by telling us that there are 803 g of fuel in 1 L, in which case:
ρ=[tex]\frac{mass}{Volume}=\frac{803g}{1L} =803\frac{g}{L}[/tex]
The required mass of fuel is 2.2 * 10⁴ kg, we can convert this value into g:
2.2 * 10⁴ kg * [tex]\frac{1000g}{1kg}[/tex] = 2.2 * 10⁷ g
We calculate the required volume (V), using the mass and density:
[tex]803 g/L = \frac{2.2*10^{7}g }{V} \\V=\frac{2.2*10^{7}g }{803g/L}\\ V=27397.26 L[/tex]
Thus 27,397.23 L would be needed for a successful trip.
