Orthoboric acid, on heating, decomposes in two ways:
(I) $H_{3} B O_{3} \to H B O_{2} + H_{2} O$
(II) $H_{5} B O_{3} \to B_{2} O_{3} + H_{2} O$
If 9 moles of $H_{3} B O_{3}$ decomposes by (I) pathway and remaining by (II) pathway. If total 11 moles of water are formed, then the number of moles of $B_{2} O_{3}$ formed is:

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Solution

The correct option is D 2
$\begin{matrix} H_{3} B O_{3} \to H B O_{2} + H_{2} O (p a t h w a y - I) H_{3} B O_{3} \to B_{2} O_{3} + H_{2} O (p a t h w a y - I I) sin c e, 9 H_{3} B O_{3} i s d e c o m p o s e d b y p a t h w a y - I . S o, t o t a l 9 m o l e s o f H_{2} O f o r m e d b y t h i s m a t h o d . a n d, T h e r e m a i n i n g (11 - 9 = 2) m o l e s o f H_{2} O o b t a i n e d b y p a t h w a r - I I s o, sin c e, p e e r m o l e o f H_{2} O o b t a i n e d w i t h! m o l e o f B_{2} O_{3} S o, w i t h 2 m o l e o f H_{2} O . w e g e t 2 m o l e o f B_{2} O_{3} b y p a t h w a y - I I H e n c e, t h e o p t i o n D i s t h e c o r r e c t a n s w e r . \end{matrix}$

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