Suppose you digested 0.25 g of a copper ore samplein 15 mL of solution. One mL of this solution is transferred to a 10 mL volumetric flask, which is filled to the calibration level. The resulting concentrationwas 0.0087M Cu2+. What is the concentration of copper in the ore sample in units of percentage?

Explanation:

10 mL = 10⁻² L

1 mL of solution contains .0087 x 10⁻² mole of Cu

15 mL of solution will contain .1305 x 10⁻² mole of Cu

.1305 x 10⁻² mole of copper = .1305 x 63.5 x 10⁻² gm of Cu

= .08286 gm

concentration of copper in the ore sample

= .08286 x 100 / .25

= 33 % approx .

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