Q.

20 ml of pure acetic acid (density =0.75 g mm1 ) is mixed with 50 gm of water (density =1 g mml1 ) at a certain temperature. Calculate the molality of acetic acid in the final solution.

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answer is 5.

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Detailed Solution

Mass of acetic acid of 20 ml = 0.75×20y

moles of CH3COOH =0.75×2060

mas of Water = 50 g

Molarity =Moles of CH3COOHMass of water (kg)

 

m=20×0.756050×1000

= 5 m

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