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Molarity problem!!!
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How many liters of 4.00M solution can be made using 100 grams of lithium bromide?
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
use the molarity equation for that last step, \(M=\dfrac{n}{L}\)
|dw:1396553514468:dw| so what next?
solve for L
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