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Molality

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Question

Calculate the molarity and mole fraction of Acetone in 2.28 molal(molality) solution of acetone in ethanol
(given: density of ethanol=0.789g/ml and density of acetone=0.788g/ml)

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Solution

$T h e r e l a t i o n b e t w e e n t h e m o l a r i t y a n d m o l a l i t y i s m = \frac{M \times 1000}{1000 d - M M_{2}} m - m o l a l i t y M - m o l a r i t y d - d e n s i t y o f t h e s o l u t i o n M_{2} - m o l a r m a s s o f t h e s o l u t e - C H_{3} C O C H_{3} = 36 + 6 + 16 = 58 m = \frac{M \times 1000}{1000 d - M M_{2}} 2.28 = \frac{M \times 1000}{1000 \times 0.789 - M \times 58} S o l v e f o r M M = 1.5888 M = 1.59 M T h e r e l a t i o n b e t w e e n m o l a l i t y a n d m o l e f r a c t i o n i s X_{2} = \frac{m M_{1}}{1 + m M_{1}} W h e r e X_{2} - m o l e f r a c t i o n o f s o l u t e M_{1} - M o l e c u l a r m a s s o f t h e s o l v e n t - e t h a n o l - C_{2} H_{5} O H = 46 m - m o l a l i t y X_{2} = \frac{m M_{1}}{1 + m M_{1}} = \frac{2.28 \times 46}{1 + 2.28 \times 46} = 0.9905 X_{1} + X_{2} = 1 (w h e r e X_{1} - m o l e f r a c t i o n o f s o l v e n t) X_{1} = 1 - X_{2} = 1 - 0.9905 = 0.00944$

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