Question

Two Samples, weighing 0.1024g and 0.9991g, are obtained from an unkown copper ore, ttreated with acid, dissolved, and each individually diluted to 100mL. The solutions are each subjected to electrolysis to remove the copper from the solution. The cathodes used in the electrolysis of each soultion, respectively, had initial masses of 13.057g and 13.084g. The final masses of the cathodes after the electrolysis were 13.141g and 13.935g.

A) What is the average % by mass of the copper in the ore?

B) What is the average molarity of copper in the 100mL dilution?

*Please provide steps*

Answer #1

Let sample 1 be the one with weight 0.1024 g and sample 2 be the one with weight 0.9991 g. We can assume that copper gets entirely deposited at the cathodes during electrolysis.

Now amount of copper deposited at the cathodes can be calculated as mass of copper = Final weight - initial weight

Mass of copper in sample 1 = 13.141 - 13.057 = 0.084 g

Mass of copper in sample 2 = 13.935 - 13.084 = 0.851 g

Now, % by mass of copper in sample 1 = (0.084/0.1024) * 100 = 0.86 % % by mass of copper in sample 2 = (0.851/0.9991) * 100 = 85 %

Therefore avg % by mass = 85.86/2 = 43.015 %

Atomic weight of copper = 63.5 g/gmol. Therefore no of moles of copper in sample 1 = 0.084/63.5 = 0.001323 gmol. No of moles of copper in sample 2 = 0.851 / 63.5 = 0.0134 gmol.

Molarity of sample 1 = 0.001323 / 0.1 = 0.01323 M

Molarity if sample 2 = 0.0134/0.1 = 0.134 M

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