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OpenStudy (anonymous):
more Chem. Calculate the amount of solute needed to prepare the following solutions: a. 50.0mL of a 5.0% (m/v) KCl solution b. 1250 mL of a 4.0% (m/v) NH4Cl solution
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OpenStudy (anonymous):
A. 0.05 = kg KCl / 0.05L B. 0.04 = kg NH4Cl / 1.25L
OpenStudy (anonymous):
ok so (%(m/v))=5.0% 100(m/50.0ml)=5.0% lets get rid of the % sign by dropping the % sign from the other side m/50.0ml=5 m=250 and i bet its in grams okf KCl
OpenStudy (anonymous):
^i meant the 100 from the other side not the % sign twice ;)
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