Calculate each of the following quantities: a) Molarity of a solution prepared by diluting 27.0 cm3
of 0.150 M potassium chloride to
150.0 cm3
b) Molarity of a solution prepared by diluting 35.71 cm3
of 0. 0756 M ammonium
sulfate to 500 cm3
c) Final volume of a 0.05M solution prepared by diluting 10.0 cm3
of 0.155 M lithium
carbonate with water

A. 0.027 M

B. 0.0054 M

C. 31 cm³

Explanation:

A. Determination of the final concentration (Molarity)

Initial Volume (V₁) = 27 cm³

Initial concentration (C₁) = 0.150 M

Final volume (V₂) = 150 cm³

Final congratulation (C₂) =?

C₁V₁ = C₂V₂

0.150 × 27 = C₂ × 150

4.05 = C₂ × 150

Divide both side by 150

C₂ = 4.05 / 150

C₂ = 0.027 M

Thus, the final concentration of the solution is 0.027 M

B. Determination of the final concentration (Molarity)

Initial Volume (V₁) = 35.71 cm³

Initial concentration (C₁) = 0.0756 M

Final volume (V₂) = 500 cm³

Final congratulation (C₂) =?

C₁V₁ = C₂V₂

0.0756 × 35.71 = C₂ × 500

Divide both side by 500

C₂ = (0.0756 × 35.71) / 500

C₂ = 0.0054 M

Thus, the final concentration of the solution is 0.0054 M

C. Determination of the final volume.

Initial Volume (V₁) = 10 cm³

Initial concentration (C₁) = 0.155 M

Final congratulation (C₂) = 0.05 M

Final volume (V₂) =?

C₁V₁ = C₂V₂

0.155 × 10 = 0.05 × V₂

1.55 = 0.05 × V₂

Divide both side by 0.05

V₂ = 1.55 / 0.05

V₂ = 31 cm³

Thus, the final volume of the solution is 31 cm³

d is

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