Ask your own question, for FREE!
Chemistry 11 Online
OpenStudy (anonymous):

A student prepares dilutions according to the following: a) 50 mL of a 6.0 M NaOH solution diluted to a final volume of 500 mL b) 75 mL of solution A are diluted to 200 mL and c) 150 mL of solution B and diluted to 1000 mL. What is the concentration of solution C?

OpenStudy (anonymous):

a) (50)(6)=500(x) solve for x b)(75)(M of solution A)=(200)(x)..are you given the concentrations?

OpenStudy (anonymous):

no. I am not given the concentrations.

OpenStudy (anonymous):

is this all there is to this question?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

is solution c a combination of all these substances?

OpenStudy (anonymous):

most likely.

OpenStudy (anonymous):

was this like a lab or something? sorry i'm trying to figure it out

OpenStudy (anonymous):

Yes, it's an example in the post lab questions and I can't seem to figure it out. Neither can my lab partner.

OpenStudy (anonymous):

do you have lab data for solution a,b, or c?

OpenStudy (anonymous):

These lab questions don't have anything to do with the physical lab we did. It just says that 75 mL of solution A was diluted to 200 mL and 150 mL of solution B was diluted to 1000 mL.

OpenStudy (anonymous):

weird...well sorry i dont know =(

OpenStudy (anonymous):

would it help if there were multiple choice?

OpenStudy (anonymous):

possible, cuz i might be able to figure out what they are asking for

OpenStudy (anonymous):

A. 0.60 M B. 0.15 M C.0.30 M D. 0.0338 M E. 0.034 M

OpenStudy (anonymous):

well i lied..that didn't help =s...the only thing i'm getting is that the total volume is 1700mL...unless..do you have moles of either A or B..i'm guessing not but just wondering

OpenStudy (anonymous):

Nope, I don't have moles of either one. The second part of the question asks: How many moles of NaOH are present in 850 mL of solution C in part a? And that too is multiple choice.

OpenStudy (anonymous):

this would be simple if you had the molarity or moles of A and B..but i really can't see a way of solving it..im sorry

OpenStudy (anonymous):

It's quite alright. I'm stuck at the exact same place you are.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!