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...

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Chemistry

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 â€“ Initialconcentration
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 16.49 mL â€“ 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.

Answer & Explanation Solved by verified expert

4.5 Ratings (946 Votes)

For acid unknown 4 weight 0118g volume 50ml For Base NaOH weight molar mass of NaOH40 volume used1149ml concentrationM 0103M For titration Macid X Vacid Mbase X Vbase Therefore See Answer

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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

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