Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 g \( mL^{-1}\)and the mass per cent of nitric acid in it being 69%.

Asked by Sudhanshu | 1 year ago |  77

1 Answer

Solution :-

Mass percent of HNO3 in sample is 69 %

Thus, 100 g of HNO3 contains 69 g of HNO3 by mass.

Molar mass of HNO3

= { 1 + 14 + 3(16)} \(g.mol^{-1}\)

= 1 + 14 + 48

\(= 63g\;mol^{-1}\)

Now, No. of moles in 69 g of \(HNO_{3}\)

\(\dfrac{69\:g}{63\:g\:mol^{-1}}\)

= 1.095 mol

Volume of 100g HNO3 solution

\(\dfrac{Mass\;of\;solution}{density\;of\;solution}\)

\(\dfrac{100g}{1.41g\;mL^{-1}}\)

= 70.92mL

\(70.92\times 10^{-3}\;L\)

Concentration of HNO3

=\(\dfrac{1.095\:mole}{70.92\times 10^{-3}L}\)

= 15.44mol/L

Therefore, Concentration of HNO3 = 15.44 mol/L

Answered by Sudhanshu | 1 year ago

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