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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

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