The density of solid sand (without air spaces) is about 2.84 g/mL. The density of gold is 19.3 g/mL.?

Question

If a 1.00 kg sample of sand containing some gold has a density of 3.10 g/mL (without air spaces), what is the percentage of gold in the sample?

Answer 1

Percent by mass?  (Not by volume?)  BTW the "1 kg" has no bearing on anything!

Let v be the volume fraction of gold in the sample.  Then

v*19.3 g/mL + (1 - v)*2.84 g/mL = 3.10 g/mL =>

2.84 g/mL + 16.46*v g/mL = 3.10 g/mL =>

v = (0.26 g/mL)/(16.46 g/mL) = 0.0158.

The MASS fraction of gold is 

[(0.0158)*19.3] / [0.0158*19.3 + 0.9842*2.84] = 9.8 %