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What mass in grams of solute is needed to prepare 900mL of 6.10×10−2M KMnO4

Chemistry
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What mass in grams of solute is needed to prepare 900

mL of 6.10×10−2M KMnO4
Oct 31st, 2014

The units of molarity are mol/L, so first we need to convert 900 mL to liters (using the conversion that 1000 mL = 1L).  Next, multiply by the molarity (6.1 x 10-2 mol/L) to give the amount of moles of KMnO4 needed.  Finally, multiply by the molecular weight of KMnO4 (158.034 g/mol). 

900 mL (1 L / 1000 mL) (6.1 x 10-2 mol/L)(158.034 g / mol) = 8.67 g KMnO4

A quick dimensional analysis shows that all the units cancel except for g of KMnO4

 

Oct 31st, 2014

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