Question

A solution is prepared by dissolving 25.00 g of acetic acid (CH3COOH) in 750.0 g of water. The density of the resulting solution is 1.105 g/mL.

A) Calculate the mass percent of acetic acid in the solution.

B) Calculate the molarity of the solution.

C) Calculate the molality of the solution.

D) Calculate the mole fraction of acetic acid in the solution.

E) What is the concentration of acetic acid in ppm?

Answer 1

a)

m/m% = mass of acit / total mass * 100% = (25)/(25+750) * 100 = 3.23%

b)

molarit = mol/V

total mass = 775 g --> V = m/D = 775/1.105 = 701.3 mL = 0.7013 L

mol = mass/MW = 25/60.05 = 0.4163 mol

so

M = mol/V = (0.4163)/0.7013 = 0.5936 M

c)

m = mol of solute / kg solvent = 0.4163 / 0.75 = 0.555 mol of solute

d)

mol frac..

mol of watr = mass/MW = 750/18 = 41.66 mol

mol of solute = 0.4163

total mol = 0.4163 +41.66 = 42.0763

x-acid = 0.4163/42.0763 = 0.009893

e)

C in PPM:

PPM = mg/kg

ppm = 25*10^3 / (750*10^-3) = 33333.3333 ppm