(A) Given: Molarity $(M) = 2.05 \ M$,Density $(d) = 1.02 \ g/mL$,Molar mass of acetic acid $(CH_3COOH) = 60 \ g/mol$. Molality $(m)$ is calculated usi

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(A) Given: Molarity $(M) = 2.05 \ M$,Density $(d) = 1.02 \ g/mL$,Molar mass of acetic acid $(CH_3COOH) = 60 \ g/mol$. Molality $(m)$ is calculated using the formula: $m = \frac{1000 \times M}{1000 \times d - M \times M_{solute}}$. Substituting the values: $m = \frac{1000 \times 2.05}{1000 \times 1.02 - 2.05 \times 60}$. $m = \frac{2050}{1020 - 123} = \frac{2050}{897} \approx 2.28 \ mol \ kg^{-1}$.

Class 12 Chemistry Solutions Method of expressing concentration of solution

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Density of a $2.05 \ M$ solution of acetic acid in water is $1.02 \ g/mL$. The molality of the solution is ............. $mol \ kg^{-1}$.

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A
$2.28$
B
$0.44$
C
$1.14$
D
$3.28$

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Density of a $2.05 \ M$ solution of acetic acid in water is $1.02 \ g/mL$. The molality of the solution is ............. $mol \ kg^{-1}$.

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