a) Calculate the w/w% of calcium chloride (CaCl2) in an aqueous solution that is

Question

a) Calculate the w/w% of calcium chloride (CaCl2) in an aqueous solution that is 2.00 M of calcium chloride and has a density of 1.17 g/mL.

b) w/v% of CaCl2 for the above solution

c) the molarity of CaCl2 in a diluted solution prepared by adding 25 mL of the above CaCl2 solution in sufficient water to give a total volume of 10L.

d) pH of the dilute CaCl2 solution

e)the ppm concentration of chloride ions (Cl-) in the above DILUTE CaCl2 solution.

The correct answers come out to a) 18.97% b) 22.2% C) .005M d) pH=7 e) 354.5 ppm, but I don't know how to get that answer so please show the steps to get there. thank you

Explanation / Answer

a) density= 1.17 g/mL molecular weight= 111g/mol

2.0 M solution means,

1000 mL = 1000x1.17g = 1170 g of the solution contains = 2.0 moles = 2x111g = 222g CaCl2

so, weight percentage = (222/1170)x100 = 0.1897x100 = 18.97%

b) 100mL of the sol contains = 222/1000x100 = 22.2 g of CaCL2

w/v% of CaCl2 for the above solution = 22.2%

c)V1S1 = V2S2

25mLx2M = 10x1000mLxS2

S2 = the molarity = 5x10^-3 M = 0.005 M

d) CaCl2 is a neutral salt [strong acid, strong base titration]

so, [H+] = 1x10^-7 and pH = -log[H+] = 7

e) from 1 CaCl2 , you get to Cl-

concentration 0f Cl-in ppm = [0.005X35.5X2] = 0.355 g/L = 355 ppm

[if you consider the atomic mass of Cl = 35.45, then the answer is 354.5]

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