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Ways of Representing the Concentration of a Solution

On a certain ...

Question

On a certain hill station pure water is found to boil at 95⁰C. How many grams of NaClis required so that water boils at 100⁰C

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Solution

$B o i l i n g p o i n t o f p u r e w a t e r ({T^{o}}_{b}) = 95^{o} C B o i l i n g p o i n t o f t h e s o l u t i o n (T_{b}) = 100^{o} C E l e v a t i o n i n b o i l i n g p o i n t (Δ T_{b}) = T_{b} - {T^{o}}_{b} = 100 - 95 = 5^{o} C Δ T_{b} = i K_{b} m i = V a n' t H o f f f a c t o r = f o r s t r o n g e l e c t r o l y t e i t i s e q u a l t o t h e n u m b e r o f i o n s i t b r e a k s i n t o i . e . f o r N a C l = 2 m = m o l a l i t y = m o l e s o f s o l u t e i n 1000 g o f s o l v e n t K_{b} = m o l a l e l e v a t i o n c o n s \tan t = 0.512^{o} C / m o l a l L e t m a s s o f w a t e r = 100 g H e n c e, 5 = 2 \times 0.512 \times \frac{W}{M} \times \frac{1000}{m a s s o f s o l v e n t (g)} W = m a s s o f N a C l M = M o l a r m a s s o f N a C l = 23 + 35.5 = 58.5 g 5 = 2 \times 0.512 \times \frac{W}{58.5} \times \frac{1000}{1000} W = \frac{5 \times 58.5 \times 1000}{0.512 \times 1000 \times 2} = 285.6 g$

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