Question

A sodium hydroxide solution is made by mixing 13.64 g NaOH with 100 g of water. The resulting solution has a density of 1.131 g/mL.

a. What is the molarity of this solution?

b. What is the molality of the solution?

c. What is the mass fraction of NaOH in the solution?

d. What is the mole fraction of NaOH in the solution?

Answer #1

Molarity = moles of solute/ L of solution

Mass of solution =mass of NaOH + mass of water= 13.64+100= 113.64 gms

Density= 1.131g/ml volume of solution = mass of soluttion/density= 113.64/1.131=100.4775 ml=0.100477 L

Molarity= moles of solute./ liter of solution

Moles of NaOH= mass/Molecular weight Since molecular weight of NaOH= 40

Moles of NaOH= 13.64/40 =0.341 moles

Molarity= 0.341/0.100477 =3.393 M

Molality =moles of solute/ kg of solvent

mass of solvent= 100g =0.1 kg

molality= 0.341/0.1=3.41 molal

mass fraction of NaOH = mass of NaOH/ total mass= 13.64/113.64= 0,1200 28

Total moles = moles of NaOH (0.341)+ moles of water (100/18)= 5.8965

Mole fraction= moles of NaOH/ total moes= 0.341/5.8965 =0.05783

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