Weight of Unknown #4( in g) | Volume of water (in mL) | Unknown #4 solution (in g/L) |
0.118 | 50.00 | 2.36 |
0.127 | 50.00 | 2.54 |
0.112 | 50.00 | 2.24 |
Volume of NaOH used = Final concentration – Initialconcentration
                                        = 16.49 mL – 4.90 mL = 11.49mL NaOH=0.01149 L NaOH
Exp. # | Concentration of NaOH | Initial volume of NaOH | Final volume of NaOH | Volume of NaOH used for titration |
1 | 0.1030M | 4.90mL | 16.49mL | 11.59mL |
2 | 0.1030M | 16.49mL | 26.01mL | 9.52mL |
3 | 0.1030M | 26.01mL | 35.09mL | 9.08mL |
a) Propose whether the unknown #4 is monoprotic or diprotic.
b) Propose the molar mass of the unknown #4.
c) show the calculation for the assumption for the monoprotic ordiprotic.
