0.5g of h2so4 oleum is diluted with water. This solution is completely neutralised by 26.7 ml of 0.4 n naoh %of free so3 in sample is?

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Solution

Dear Student,

The reaction scheme is,

$S O_{3} + 2 N a O H \to N a_{2} S O_{4} + H_{2} O 1 m o l e o f S O_{3} r e a c t s w i t h 2 m o l e o f O H^{-} i o n s, E q u i v a l e n t m a s s o f S O_{3} = \frac{M o l e c u l a r m a s s o f S O_{3}}{2} = \frac{32 + 3 (16)}{2} = \frac{80}{2} = 40 \frac{g}{e q u i v} \sin c e 2 H^{+} i o n s r e a c t s p e r H_{2} S O_{4} m o l e c u l e, E q u i v a l e n t m a s s o f H_{2} S O_{4} = \frac{M o l e c u l a r m a s s o f H_{2} S O_{4}}{2} = \frac{2 (1) + 32 + 4 (16)}{2} = \frac{98}{2} = 49 \frac{g}{e q u i v} L e t' s c o n s i d e r x g m o f S O_{3} p r e s e n t i n 0.5 g m o l e u m M a s s s u l f u r i c a c i d i n 0.5 g m o l e u m = (0.5 - x) g m E q u i v a l e n t o f H_{2} S O_{4} = \frac{0.5 - x}{49} E q u i v a l e n t o f S O_{3} = \frac{x}{40} E q u i v a l e n t o f N a O H = 0.4 N \times (\frac{26.7}{1000}) l i t r e s = 0.01068 e q u i v . W e k n o w 1 g m - e q u i v o f a c i d s n e u t r a l i z e s 1 g m - e q u i v o f b a s e s H e n c e w e g e t, \frac{x}{40} + \frac{(0.5 - x)}{49} = 0.01068 o n s o l v i n g w e g e t, 49 x + 40 (0.5) - 40 x = 0.01068 (49) (40) 9 x = 0.9328 g m x = 0.103 g m L e t' s c o m p u t e f o r %, % o f f r e e S O_{3} = \frac{M a s s o f S O_{3}}{T o t a l m a s s} \times 100 = \frac{0.103 g m}{0.5 g m} \times 100 % = 20.6 %$

Regards.

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