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Chemistry / Chemistry 11 / Chapter 1 / Problem 93P

Chemistry | 11th Edition | ISBN: 9780073402680 | Authors: Raymond Chang

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Solved: In 1849 a gold prospector in California collected

Chapter 1, Problem 93P

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QUESTION:

In 1849 a gold prospector in California collected a bag of gold nuggets plus sand. Given that the density of gold and sand are \(19.3\mathrm{\ g}/\mathrm{cm}^3\) and \(2.95\ \mathrm{g}/\mathrm{cm}^3\) respectively, and that the density of the mixture is \(4.17\mathrm{\ g}/\mathrm{cm}^3\). calculate the percent by mass of gold in the mixture.

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QUESTION:

In 1849 a gold prospector in California collected a bag of gold nuggets plus sand. Given that the density of gold and sand are \(19.3\mathrm{\ g}/\mathrm{cm}^3\) and \(2.95\ \mathrm{g}/\mathrm{cm}^3\) respectively, and that the density of the mixture is \(4.17\mathrm{\ g}/\mathrm{cm}^3\). calculate the percent by mass of gold in the mixture.

ANSWER:

Step 1 of 4

Here, we are going to calculate the percent gold in mass from the mixture.

Given that,

Density of Gold () = 19.3 g/

Density of sand ()= 2.95 g/

Density of the mixture () = 4.17 g/

We know that,

Density is defined as the mass per unit volume.

$Density\ =Mass/volume$

$Mass\ =Density\ \times{}\ volume.$

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