Three samples of an unknown monoprotic acid were titrated with standardized NaOH

Question

Three samples of an unknown monoprotic acid were titrated with standardized NaOH and give calculated molar masses of 88.8, 90.3 and 89.0 g/mol. What is the average molar mass (in g/mol)? Tries 0/3 What is the relative standard deviation(RSD) in parts per thousand (ppt)? Report your answer to the correct number of significant figures. For help with RSD calculations see the techniques section in your lab manual ppt Tries 0/3 If RSD is> 5ppt, discard the MM furthest from the average and recalculate the average MM and RSD. What is the new average MM (in g/mol)? Tries 0/3 What is the new RSD (in ppt)? Report your answer to the correct number of significant figures ppt Tries 0/3

Explanation / Answer

Q1

Average molar mass = sum of masses / total samples

Avg. = (88.8+90.3+89)/3

Avg. = 89.36666 g/mol

Q2

find RSD

RSD = stedv/avg

stdev = sqrt(( 88.8-89.36666)^2 + (90.3-89.36666 )^2 + (89.00-89.36666)^2)/3

stdev = 0.38393

avg = 89.36666

RSD ppt = stedev/avg * 1000 = 0.38393/89.36666 *1000 = 4.29612

note that RSD < 5

so keep it

no need to reclacualte values

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