Posted by Anonymous on .
How many grams of K2Cr2O7 are required to prepare 500 ml of a 0.13 N solution?
Analytical chemisty -
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