Question

1.00 mL of a 0.0500 M sucrose solution is transferred to a 100 mL volumetric flask, and the flask brought up to volume. 5.00 mL are taken from the 100 mL volumetric flask and transferred to a 0.500 L flask, and it is brought up to volume. What is the resulting concentration of sucrose in the 0.500 L flask?

Answer 1

you diluted it by a factor of 100 (added 99 parts water). then, you diluted it another factor of 100 (.005L to .5L).

New concentration: .0500/1E6 M

Answer 2

i don't understand, can u explain using the
formula m1v1=m2v2?

Answer 3

Not with that formula.

Answer 4

oh, is there an easier way to this problem?

Answer 5

I erred. Goodness. You diluted it 100 times, then diluted it 100 times again.

new concentration=.05M/(100x100)=.05/10,000 or .05*10^-4 or 5*10^-6 M

Answer 6

If you want to do it by the m1v1 = m2v2 formula you can do it twice.
The first one is 1x.05 = 100*m
m = 0.05/100 = 5E-4
Then m1v1 = m2v2 the second time.
5E-4*5mL = 500mL x M
5E-4 x (5/500) = 5E-6M
If you insist on a formula that will do it; however, I like to do it this way.
0.05 M is what we start with. So
0.05M x (1/100) x (5/500) = %E-6M

Answer 7

oops. 5E-6

Answer 8

thanks drbob22, much better explanation!

Answer 9

ccny?