The density of a $2.0 M$ solution of acetic acid in water is $1.01 g {cm}^{- 3}$ . The molality of acetic acid in solution is

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$2.12 mol {kg}^{- 1}$

$2.50 mol {kg}^{- 1}$

$3.0 mol {kg}^{- 1}$

$2.25 mol {kg}^{- 1}$

answer is B.

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Detailed Solution

For 2.0 solution, we will have

Amount of acetic acid, $n_{2} = 2.0 mol$ Volume of solution, $V = 1000 {cm}^{3}$

Mass of solution $m = ρV = (1 .01 g {cm}^{- 3}) (1000 {cm}^{3}) = 1010 g$

Mass of acetic acid, $m_{2} = n_{2} M_{2} = (2 .0 mol) (60 g {mol}^{- 1}) = 120 g$

Mass of water in $1010 g$ of solution $m_{1} = 1010 g - 120 g = 890 g = 0 .890 kg$

Molality of acetic acid $= \frac{n_{2}}{m_{1}} = \frac{2 .0 mol}{0 .890 kg} = 2 .25 molkg$

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