Mole fraction of solvent in aqueous solution of naoh having molality of 3 is ?

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Solution

Dear Student,

The computation are as expressed below,

$3 m o f s o l u t i o n d e n o t e s 3 m o l e s o f s o l u t e i n 1000 g m o f s o l v e n t (w a t e r) N o . o f m o l e s o f N a O H = 3 N o . o f m o l e s o f w a t e r (s o l v e n t) = \frac{m a s s}{m o l a r m a s s} = \frac{1000}{18} = 55.55 m o l e s M o l e f r a c t i o n = \frac{m o l e s o f s o l v e n t}{m o l e s o f s o l u t e + m o l e s o f s o l v e n t} = \frac{55.55}{3 + 55.55} = \frac{55.55}{58.55} = 0.94876 \approx 0.95$
Regards.

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