Question

Consider that you have 200.mL of 0.0500 M Na2SO4 solution in water. The density of the solution is 1.05 g/mL.

a) What os the mass % of sodium sulfate in the solution?

b) What is the ppm of the sodium sulfate in the solution?

c) What is the molarity of ions in the solution assuming 100% dissociation?

d) What is the molality of sodium sulfae in the solution?

e) What is the molality of ions in the solution assuming 100% dissociation?

Answer 1

(a)

Moles of solute = Molarity*Volume = 0.05*0.2 = 0.01

Mass of solute = Moles*MW = 0.01*142 = 1.42 g

So,

Mass % of solute = (Mass/Volume in mL)*100 = (1.42/200)*100 = 0.71%

(b)

Mass of solute = 1.42 g = 1420 mg

Volume = 200 mL = 0.2 L

So,

Conc in ppm = Conc in mg/L = 1420/0.2 = 7100 ppm

(c)

Molarity of sodium ions = 2*0.05 = 0.1 M

Molarity of sulfate ions = 0.05 M

(d)

Molality of solute = Moles/Mass of solvent in kgs = Moles/(Density*Volume) = 0.01/((200*1.05)/1000) = 0.0476 molal

Hope this helps !