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3 years ago

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What is the density of 18.0-karat gold that is a mixture of 18 parts gold, 5 parts silver, and 1 part copper? (These values are

parts by mass, not volume). Assume that this is a simple mixture having an average density equal to the weighted densities of its constituents. The density of gold is 19.32g/cm3, silver is 10.1g/cm3 and copper is 8.8g/cm3

1 answer:

nexus9112 [7]3 years ago

8 0

Answer:

Density of 18.0-karat gold mixture is $15.58 g/cm^3$ .

Explanation:

A mixture of 18 parts gold, 5 parts silver, and 1 part copper.

Let mass of gold be 18x

Let the mass of silver be 5x

Let the mass of copper be 1x

The density of gold = $19.32g/cm^3$

The density of silver = $10.1g/cm^3$

The density of copper = $8.8g/cm^3$

$Volume=\frac{Mass}{Density}$

Volume of the gold in the mixture = $V_1=\frac{18x}{19.32 g/cm^3}$

Volume of the silver in the mixture = $V_2=\frac{5x}{10.1 g/cm^3}$

Volume of the copper in the mixture = $V_3=\frac{1x}{8.8 g/cm^3}$

Mass of the mixture = M = 18x+5x+1x =24x

Volume of the mixture = $V_1+V_2+V_3$

Density of the mixture:

$\frac{M}{V_1+V_2+V_3}=15.58 g/cm^3$

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