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OpenStudy (anonymous):

how many liters of 4 M SOLUTION CAN BE MADE USING 100 GRAMS OF LITHIUM BROMIDE??

OpenStudy (anonymous):

n=cv where: n=amount of substance (mol) c=concentration (M) v=volume (litres) and n=m/M where: n=amount of substance (mol) m=mass (g) M=molar mass (g/mol)

OpenStudy (anonymous):

LiBr n=m/M n=100/(6.9+79.9) n=1.15 mol n=cv 1.15=4v v=0.288L \[\therefore\]you can make 0.288 L of 4M solution using 100g of lithium bromide

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