How many number of active hydrogen atoms are present in a compound(Mol. mass 90) 0.45 g of which when treated with sodium metal liberates 112 ml of the hydrogen gas at STP?

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Solution

$S i n c e h y d r o g e n i s l i b e r a t e d, t h e c o m p o u n d m a y b e a l c o h o l o r a c i d . N a + R C O O H \to R C O O N a + \frac{1}{2} H_{2} 0.45 g o f s o d i u m l i b e r a t e s 112 m l h y d r o g e n 23 g r e a c t s w i t h a c i d o f m o l e c u l a r m a s s 90 = \frac{112}{0.45} \times 23 = 5724.4 m l A t s t p 22400 m l i s t h e v o l u m e o c c u p i e d b y 1 m o l e o f H_{2} s o 5724.4 m l = \frac{1}{22400} \times 5724.4 = 0.25 m o l e o f H_{2} N o o f h y d r o g e n a t o m = 0.25 \times 2 \times 6.022 \times 10^{23} (1 m o l e c u l e h a s 2 a t o m s o f h y d r o g e n) = = 3.011 \times 10^{23} a t o m s$

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