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 .

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole