Posted by Anonymous on Saturday, February 2, 2013 at 7:33pm.
How many grams of K2Cr2O7 are required to prepare 500 ml of a 0.13 N solution?

Analytical chemisty  Devron, Saturday, February 2, 2013 at 10:51pm
0.13N= # of equivalents/500 x10^3 L, solving for # of equivalents
# of equivalents in 0.13N=(0.13N)(500 x 10^3L)
# of equivalents in 0.13N= g of solute/49.03g, solve for grams of solute,
(49.03g, solve )*(# of equivalents in 0.13N)= g of solute
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