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OpenStudy (anonymous):

determine the amount in grams , of sodium carbonate needed to make .50mL of .300M sodium carbonate?

OpenStudy (anonymous):

So, Molarity is equal to \[=moles/Liters\] correct? So convert your .50mL to L and cancel the units. That will leave you with moles of \[NaCO _{3}\] from there you will need to convert from molarity to grams using the molecular weight \[NaCO_{3}\]. Which will then give you the answer you're looking for. |dw:1350946696986:dw|

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