# Homework Help: Finding the volume of gold in intergalactic space

1. May 11, 2006

### SnackMan78

The average density of the material in intergalactic space is approximately 2.5 × 10^–27 kg/m^3. What is the volume of a gold sample, ρ = 19 300 kg/m^3, that has the same mass as 8.0 × 10^24/ m^3 of intergalactic space?

a. 1.0 × 10^–6 m^3
b. 2.0 × 10^–5 m^3
c. 5.0 × 10^–5 m^3
d. 8.0 × 10^–5 m^3
e. 1.0 × 10^–4 m^3

Using the formula p=m/V, (where p = mass density, m=mass, V=volume) I’ve got:

19300 kg/m = (8.0 * 10^24) / V, V= 2.41 x 10^-21.

Based on the answer options, I have more computations to do, but where? I see I haven't used the value for the avg. density(2.5 x 10^-27), but I'm being asked for volume. Does the volume change because of intergalactic space?

At this point I'm confused. Can anyone point me in the right direction to solve this question?

2. May 11, 2006

### Kurdt

Staff Emeritus
You have used 8x1024 as a mass but it is a volume.

3. May 15, 2006

### SnackMan78

Then switching values I get: 19300 / 8 x1024 = mass,
m = 2.4 x 10-21. Substituting my "m" value into the first half of the question: 2.5 x10-27 / 2.4 x 10-21 = v, v = 1.04 x 10-6. So is the answer "A"? Or have I missed interpreted your help?

Note: I don't how to get the exponents into superscript form as done in Kurdt's response.

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