6.b.
For Trial 2
The molar concentration of the acid = 0.922 M
For Trial 3
The molar concentration of the acid = 0.856 M
The average molar concentration of the acid solution = 0.922 M + 0.856M + 0.08 / 3
= 1.858/3
= 0.619 M acid solution
b.
Standard deviation = √[(0.619 – 0.922)^2 + (0.619 - 0.856)^2 + (0.619-0.08)^2]/2]
= √[(0.619)^2 –2 x 0.619 x 0.922 + (0.922)^2 + (0.619)^2 –2 x 0.619 x 0.856 + (0.856)^2 + (0.619)^2 -2 0.619 x 0.08 + (0.08)^2]/2
= √[(0.0918 + 0.0561 + 0.291) / 2]
= 0.468 M Standard Deviation
Relative Standard Deviation = (0.468 / 0.619) x 100
= 75.6 % RSD
6. b. For Trials 2 and 3, the molar concentration of the acid was 0.922 M and 0.856 M respectively. a. What is the ave...
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