Mass of Vial + Unknown Acid =16.245 Mass of Vial - 1st unknown acid sample = 16.

ID: 1089173 • Letter: M

Question

Mass of Vial + Unknown Acid =16.245

Mass of Vial - 1st unknown acid sample = 16.068

Mass of Vial - 2nd unknown acid sample = 15.876

Mass of Vial - 3rd unknown acid sample = 15.443

31.5 mL

Calculate the molar mass mass of your unknown acid using 1st, 2nd , and 3rd titration. Trial 1 .1749M Trial 2 .1959M Trial 3 .22319M

----- 1st titration 2nd titration 3rd titration Initial Vol NaOH 1mL 7mL 0mL Final Vol NaOH 26.3mL

31.5 mL

48.5mL Vol of NaOH used 25.3 mL 24.5mL 48.5 mL Mass of unknown solid .177g .192g .433g

Given:

Trial 1: Mass of unknown solid=0.177g

molar mass of unknown acid(MW)= mass/mol of unknown acid

So,mol of unknown acid is to be determined from titration using equation:

M(NaOH)*V(NaOH)=M(acid)*V(acid)=mol of acid=mol of NaOH

where

Molarity of NaOH=M(NaOH)

Molarity of acid=M(HCl)

Volume of NaOH=V(NaOH)

Volume of acid=V(acid)

trial 1)

Molarity of NaOH=M(NaOH)=0.1749M

Volume of NaOH=V(NaOH)=25.3 ml=0.0253L

mol of acid=0.1749 mol/L*(0.0253L)=0.00442mol

MW =0.177g/0.00442mol=40.045g/mol

trial 2:

Molarity of NaOH=M(NaOH)=0.1959M

Volume of NaOH=V(NaOH)=24.5 ml=0.0245L

mol of acid=0.1959 mol/L*(0.0245L)=0.00480mol

MW =0.192g/0.00480mol=40.0g/mol

trial 3:

Molarity of NaOH=M(NaOH)=0.22319M

Volume of NaOH=V(NaOH)=48.5 ml=0.0485L

mol of acid=0.22319 mol/L*(0.0485L)=0.0108mol

MW =0.433g/0.0108mol=40.092g/mol

MW(average)=(40.045+40.0+40.092)/3=40.046g/mol