What is the mass of #"10 m"^3# of water given that water has a density of #"1 g/cm"^3# ?

1 Answer
Jun 14, 2015

Answer:

The mass of that much water will be equal to #10^4"kg"#.

Explanation:

Notice that you were given a volume of water, #"10 m"^3# to be precise, and asked to determine the mass, in kilograms, this much water would have.

The only thing you need to do to determine the mass of water is to use the water's density.

Density is defined as mass per unit of volume. In water's case, you know that 1 gram of water occupies exactly #"1 cm"^3#. To get from grams to kilograms and from cubic centimeters to cubic meters, you're going to have to use a series of conversion factors.

In one single step, you would get

#10cancel("m"^3) * (10^3cancel("dm"^3))/(1cancel("m"^3)) * (10^3cancel("cm"^3))/(1cancel("dm"^3)) * (1cancel("g"))/(1cancel("cm"^3)) * (1"kg")/(10^3cancel("g")) = (10 * 10^3 * cancel(10^3)"kg")/(cancel(10^3)) = color(green)(10^4"kg")#

Another way to look at this is

#rho = m/V => m = rho * V#

Since you're given volume, you can determine the mass in grams first, then convert it to kilograms.

#m = "1 g"/(1cancel("cm"^3)) * (10^6 cancel("cm"^3))/(1cancel("m"^3)) * 10cancel("m"^3) = 10^7"g"#

Expressed in kilograms, you'd once again get

#10^7cancel("g") * "1 kg"/(10^3cancel("g")) = color(green)(10^4"kg")#