Question

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 average molar concentration of the acid solution? Data Analysis, B. b, what are the standard deviation and the relative standard deviation (%RSD) for the molar concentration of the acid solution? Data Analysis, C and D.

Answer 1

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