Asked by confused

Three samples of an unknown acid are titrated with 0.4901 M NaOH. The average equivalent weight of the acid using the data below is g/equiv
mass of unknown vol NaOH
trial 1: 1.02 27.72
trial 2: .522 14.22
trial 3: .742 20.02
Answered by DrBob222
Your question isn't clear. Please clarify it. Correct spacing on these boards is a problem (meaning you can't do it); therefore, you must explain the problem in words and not tables.
Answered by Confused
Three samples of an unknown acid are titrated with 0.4901 M NaOH. The average equivalent weight of the acid using the data below is g/equiv ?

the mass of unknown for trial 1 is 1.02, trial 2 is 0.522, trial 3 is 0.742
volume of NaOH in trial 1 is 27.72, trial 2 is 14.22, and trial 3 is 20.02
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