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 Homework Statement:
 What volume of 40% NaNO3 solution contains 0.15 mole of NaNO3 ? Density = 1.32 g/mL = 1320 g/L
 Relevant Equations:

mwt of NaNO3 = 85 0.15 moles of NaNO3 = 12.5 grams
Soluability of NaNO3 is 900 g/L at 25 C
Since the density is 1320 g/L then for every L of solution there is 320 grams of solute.
320 is 35% of 900 hence the statement it is is a (about) a 40% solution
So the volume that would have 12.5 grams would be 12.5/320 L = .039 L = 39 mL
320 is 35% of 900 hence the statement it is is a (about) a 40% solution
So the volume that would have 12.5 grams would be 12.5/320 L = .039 L = 39 mL