Question

how do i figure out the percentage error , as well as the absolute error of
a) (10.00+/- .02)mL x .42 mol/L
b) (10.00 +/- .02)mL + (10.00 +/- .02)mL

thanks in advancee plz help

Answer 1

a) The % error is 0.02/10 = 0.2%
The absolute error is 0.2% of the product. The product is 4.2*10^-3 mol Take 0.2% of that for the absolute error.

b) The errors, being supposedly random, add as the square root of the sum of the squares. The result is 20.00 +/- 0.028. That is 0.14% relative error.

Answer 2

for b where do u get the =/e .028?
and what about the absolute error?

Answer 3

I already stated the absolute error in (b).

As I said before, 0.028 is the square root of the sum of two 0.02's squared.

It is possible that your teacher wants you to add the absolute errors, and not take the square root of the sum of the squares. To do so would give you an overly conservative worst case error. Errors should be handled statistically.

Refer to
http://en.wikipedia.org/wiki/Mean_squared_error