# A swimming pool measures 50.0 meters by 25.0 meters. How many grams of water are needed to fill the pool, whose average depth is 7.8 feet? Assume the density of water to be 1.0 g/mL.

Aug 31, 2016

The mass of water needed is 3.0 × 10^9color(white)(l) "g".

#### Explanation:

The formula for density is

color(blue)(|bar(ul(color(white)(a/a) "Density" = "Mass"/"Volume" color(white)(a/a)|)))" " or color(blue)(|bar(ul(color(white)(a/a) ρ = m/Vcolor(white)(a/a)|)))" "

And the formula for volume is

color(blue)(|bar(ul(color(white)(a/a) "Volume" = "length" × "width" × "height"color(white)(a/a)|)))" " or $\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} V = l w h \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

Step 1. Convert the depth of the pool from feet to metres.

h = 7.8 color(red)(cancel(color(black)("ft"))) × (12 color(red)(cancel(color(black)("in"))))/(1 color(red)(cancel(color(black)("ft")))) × "1 m"/(39.37 color(red)(cancel(color(black)("in")))) = 2.38 "m"

Step 2. Calculate the volume of the water

$V = l w h = {\text{50.0 m× 25.0 m × 2.38 m" = 2980color(white)(l) "m}}^{3}$

Step 3. Convert the volume of the water to millilitres.

The density of water is given as 1.0 g/mL, so we should convert the volume to millilitres.

V = 2980 color(red)(cancel(color(black)("m"^3))) ×( 1000 color(red)(cancel(color(black)("L"))))/(1 color(red)(cancel(color(black)("m"^3)))) × "1000 mL"/(1 color(red)(cancel(color(black)("L")))) = 2.98 × 10^9color(white)(l) "mL"

Step 4. Calculate the mass of the water.

Since ρ = m/V, m = ρ V

$m = \text{1.0 g"/(1 color(red)(cancel(color(black)("mL")))) × 2.98 × 10^9 "mL" = 3.0 × 10^9color(white)(l) "g}$