Question

Density measurements can be used to analyze mixtures. For example, the density of solid sand (without air spaces) is about 2.84 g/mL. The density of gold is 19.3 g/mL. 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

Answer:

Following are the responses to the given question:

Explanation:

Let S = solid sandy mass inside the blend

Let G = gold mass in such a blend

We search that percent G by mass

Mixture density = total blend mass / total blend volume = 3.10 g/mL

Mixture sum $= S + G = 1 \ kg = 1000\ g$

Complete mixing volume = sand volume + gold volume volume

Complete mixing volume $= (\frac{sand\ mass}{sand \ density}) + (\frac{gold\ mass}{gold \ density})$

Total mixture volume:

$= (\frac{S}{2.84}) + (\frac{G}{19.3})\\\\ = \frac{(19.3 S + 2.84 G)}{54.812\ mL}$

$\text{Mixture Density} = \frac{\text{mass mixture}}{\text{mixture volume}}$

$= 3.10 \frac{g}{mL} \\\\ \to 3.10 = \frac{1000 \ g}{ \frac{(19.3 S + 2.84 G)}{ 54.812}}\\\\\to 19.3 S + 2.84 G = 17681$

When

$S + G = 1000 \ g$

Solving equation:

$\to 19.3 S + 2.84 G = 17 681\\\\\therefore \ S + G = 1000 \ g\\\\\to 19.3 S + 2.84 G = 17 681\\\\\to -2.84S -2.84 G = -2840\\\\----------------\\\\\to 16.46S = 14 841\\\\\to S = 901.7 g \ SanD$

When  

$\to S+ G = 1000\\\\\to G = 1000 - 901.7\ g\\\to G = 98.3 g \ GOLD$