I believe there to be an error in my textbook regarding this problem;
A sodium nitrate solution is 5% NaNO3 by mass and has a density of 1.02g/mL. Calculate the molarity of the solution.
It says the answer is 1.5M but I can’t get that. I keep getting .6M
My calculations are as follows
1000mL x 1.02g/mL= 1020 g of soln x 5% NaNO3= 51 g NaNO3 per 1 L soln.
51g/84.99g= .6 mol/1L soln= .6M
Am I messing up or is the book wrong?
2 Answers

Here’s the other way, starting with mass, not volume:
1000 g of solution contains 50 g of NaNO3
1000 g divided by 1.02 g/mL = 980.392 mL
MV = mass / molar mass
(M) (0.980392 L) = 50 g / 84.99 g/mol
M = 0.60 mol/L
“Am I messing up or is the book wrong?”
Nope. You nailed the calculation correctly. The book has a mistake. Good on ya!

5% by mass = 5g solute in 100g solution
OR 50g solute in 1000g solution
Volume of 1000g solution = mass / density
Volume = 1000/1.02 = 980.39mL
Then you have 50/980.39*1000 = 51g/L solution
Molar mass NaNO3 = 23+14+16*3 = 85g/mol
mol in 51g = 51/85 =0.6 mol NaNO3 per litre of solution
You are correct.