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Q)

Density of a 2.05M solution of acetic acid in water is 1.02 g/mL. The molality of the solution is

$\begin{array}{1 1}(a)\;0.44molkg^{-1}&(b)\;1.14molkg^{-1}\\(c)\;3.28molkg^{-1}&(d)\;2.28molkg^{-1}\end{array}$

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A)
Weight of acetic acid =$2.05\times 60=123$
Weight of the solution =$1000\times 1.02=1020$
$\therefore$ Weight of water=$(1020-123)=897g$
$\therefore$ Molality =$\large\frac{2.05\times 1000}{897}= 2.285$
Hence (d) is the correct answer.
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