# Commercial aqueous nitric acid has a density of 1.42 g/mL and is 16 M. Calculate the ercent HNO3 by mass…?

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in solution.

• Atomic weights: H = 1 N = 14 O = 16 HNO3 = 63

16molHNO3/1LHNO3(s) x 1LHNO3(s)/1000mLHNO3(s) x 63gHNO3/1molHNO3 x 1mLHNO3(s)/1.42gHNO3(s) x 100% = (16)(63)(100)/(1000)(1.42) =

In solving this by dimensional analysis, the factor label method, one has to distinguish between grams and mL of nitric acid solution HNO3(s) and nitric acid HNO3 the substance.

The first term is given from the molarity. The next factor is conversion between L and mL. The LHNO3(s) cancel, leaving molHNO3 per mLHNO3(s). The next factor comes from the molecular weight of HNO3. The molHNO3(s) cancel, leaving gHNO3 per mLHNO3(s). The next factor comes from the density. The mLHNO3(s) cancel, leaving gHNO3 per gHNO3(s). Multiplying by 100% gives the result as %, which was desired.

• Nitric Acid Density

• The molecular mass of nitric acid (HNO3) is 63.0g/mol.

A 16M (or mol/L) solution contains 16 moles of HNO3 per litre of solution, which is 16×63.0 = 1008g of HNO3 per litre.

To convert to percent by mass (% w/w) you need to know how much that one litre of nitric acid solution weighs. This is easy as you know 1 mL = 1.42g, so 1L = 1420g as there are 1000mL in one litre.

Now to calculate the weight percent you divide the weight of HNO3 by the weight of solution and multiply by 100 (the same as to calculate any %).

So (1008/1420)*100 = 71%w/w HNO3

(the stuff we use here is labelled as 68 to 72%)

• First you need to know what M(molarity) is. It is actually g/L

So 16M = 16g/L

Just multiply 16 by 1.42 g/mL but first you must convert

1.42g/L = 0.00142g/L or M

I dont have a calculator so you can finish this up

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