Answer:
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 .
