Question #67459
how many grams of potassium nitrate will you need to prepare 10L of a .75M solution
Expert's answer
СM = n/V n = m/M
M (KNO3) = 101 g/mol
n (KNO3) = СM x V = 0.75 x 10 = 7.5 mol
m (KNO3) = n x M = 7.5 x 101 = 757.5 g
M (KNO3) = 101 g/mol
n (KNO3) = СM x V = 0.75 x 10 = 7.5 mol
m (KNO3) = n x M = 7.5 x 101 = 757.5 g
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#339914 on Dec 2023
#339914 on Dec 2023