Ask your own question, for FREE!
Chemistry 16 Online
OpenStudy (anonymous):

Molarity problem!!!

OpenStudy (anonymous):

How many liters of 4.00M solution can be made using 100 grams of lithium bromide?

OpenStudy (anonymous):

I got that 110g of LiBr= 1.15mol But I can't seem to figure out how to do the math to find the number of liters

OpenStudy (aaronq):

use the molarity equation for that last step, \(M=\dfrac{n}{L}\)

OpenStudy (anonymous):

|dw:1396553514468:dw| so what next?

OpenStudy (aaronq):

solve for L

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!