Calculating Molarity from Mass %?

NetherCraft 0

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?

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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.

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