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Q.

20 ml of pure acetic acid (density $= 0.75 g {mm}^{- 1}$ ) is mixed with 50 gm of water (density $= 1 g m {ml}^{- 1}$ ) 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 = $0.75 \times 20 y$

moles of ${CH}_{3} COOH$ $= \frac{0.75 \times 20}{60}$

mas of Water = 50 g

Molarity $= \frac{Moles of {CH}_{3} COOH}{Mass of water (kg)}$

$m = [\frac{\frac{20 \times 0 .75}{60}}{50}] \times 1000$

$= 5 m$

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