Question

What is the mole fraction of a NaOH in an aqueous solution that contains 33.9% NaOH by mass?

Answer 1

If the mass of solution is 100 g, than

m(NaOH)=33.9 g;

m(H₂O)=100-33.9=66.1 g

n(NaOH)=m/Mr=33.9/40=0.8475 mol;

n(H₂O)=m/Mr=66.1/18=3.67 mol;

Mole fraction of NaOH (X_NaOH):

X_NaOH= n(NaOH)/(n(NaOH)+n(H₂O))*100%=0.8475/(0.8475+3.67)*100%=18.76%

Answer: mole fraction of NaOH is 18.76%