QUESTION 2.4

Concentrated nitric acid used in laboratory work is $68%68\%$ nitric acid by aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g $mL^{-1}$ ?

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Solution

$68 % H N O_{3}$ by mass means
Mass of nitric acid = 68 g
Mass of solution = 100 g
Molar mass of $H N O_{3} = (1) + (14) + (16 \times 3) = 63 g m o l^{- 1}$
Number of moles of $H N O_{3} (n_{H N O_{3}}) = \frac{W}{M} = \frac{(68 g)}{(63 g / m o l)} = 1.079 m o l$
Density of solution $= 1.504 g m L^{- 1}$
Volume of solution $= \frac{M a s s}{D e n s i t y} = \frac{(100 g)}{(1.504 g m L^{- 1})}$
=66.5 mL or 0.0665 L
Molarity $= \frac{M o l e s o f s o l u t e}{V o l u m e o f s o l u t i o n i n l i t r e s}$
$= \frac{(1.079 m o l)}{(0.0665 L)}$
$= 16.23 m o l L^{- 1}$ or $16.23 M$

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QUESTION 2.4 Concentrated nitric acid used in laboratory work is 68%68\%68% nitric acid by aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mL−1mL^{-1}mL−1 ?

QUESTION 2.4

Concentrated nitric acid used in laboratory work is $68%68\%$ nitric acid by aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g $mL^{-1}$ ?