Determine the volume of dilute nitric acid ( $d = 1.11$ g $m L^{- 1}$ , $19 %$ w/w $H N O_{3}$ ) that can be prepared by diluting with water $50$ mL of conc. $H N O_{3}$ ( $d = 1.42$ g $m L^{- 1}$ , $69.8 %$ w/w).

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Solution

According to Volumetric Principle: $N_{1} V_{1} = N_{2} V_{2}$
Let $N_{1}$ and $V_{1}$ be Normality and Volume of dilute nitric acid respectively. Similarly, $N_{2}$ and $V_{2}$ be Normality and Volume of conc. $H N O_{3}$ respectively.

$\frac{N o . o f g - e q u i v a l e n t o f d i l . n i t r i c a c i d}{V o l u m e o f s o l u t i o n i n L} \times V_{1} = \frac{N o . o f g - e q u i v a l e n t o f c o n c . n i t r i c a c i d}{V o l u m e o f s o l u t i o n i n L} \times V_{2}$

$\frac{\frac{g i v e n m a s s}{m o l e c u l a r m a s s}}{V o l u m e o f s o l u t i o n i n L} \times V_{1} = \frac{\frac{g i v e n m a s s}{m o l e c u l a r m a s s}}{V o l u m e o f s o l u t i o n i n L} \times V_{2}$

Volume of solution $= 100 m L = \frac{100 L}{1000} = 0.1 L$

$\frac{\frac{19}{36.5}}{0.1} \times V_{1} = \frac{\frac{69.8}{36.5}}{0.1} \times 50$

$V_{1} = \frac{50 \times \frac{69.8 / 36.5}{0.1}}{\frac{19 / 36.5}{0.1}} = 183.69$ ml

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