Question

PLEASE SHOW ALL WORK WHERE RELEVANT

18. A solution is prepared by dissolving 12.36 g of CaCl2 (an electrolyte, molar mass = 111 g) in 135g of H2O (molar mass
= 18.0 g). The resulting solution has a density of 1.10 g/mL. Calculate the molarity of CaCl2 in the solution.

22. The greater the __________ , the better the substance is at forming a solution that can conduct an electrical current.
A. molecular mass

B. dielectric constant
C. degree of ionization

D. both a and b
E. both b and c

19. Consider a 0.700 M Al2(SO4)3 solution. This solution has a sulfate ion concentration of _____ .

29. Which of the following is not a colloid?
A. fog

B. air

C. mayonnaise

D. a gel

E. milk

Answer 1

18. A solution is prepared by dissolving 12.36 g of CaCl₂ (an electrolyte, molar mass = 111 g) in 135g of H₂O (molar mass = 18.0 g). The resulting solution has a density of 1.10 g/mL. Calculate the molarity of CaCl2 in the solution.

1 mole of CaCl₂ in 1000 mL of H₂O =1 M

111 g of CaCl₂ in 1000 mL of H₂O =1 M

12.36 g of CaCl₂ in 1000 mL of H₂O = (1 x 12.36) /111 M = 1.1 x 10^-1 M

  12.36 g of CaCl₂ in 135g mL of H₂O = (0.11 x 135) /1000 M = 1.48 x 10^-2 M

Now, solution has a density of 1.10 g/mL

So, the molarity of CaCl2 in the solution = (1.48 x 10^-2 ) / 1.1 M = 1.3 x 10^-2 M.

22. E. both b and c

19.

From the formula 1 mole Al₂(SO4)₃ contains 3 moles SO₄^2- and 2 moles Al³⁺

Therefore, Sulphate ion concentration is 3 x 0.7 = 2.1 M

29. B. air