Density of $2.03 M$ aqueous solution of acetic acid is $1.017 g {m L}^{- 1}$ . Molecular mass of acetic acid is $60$ . Calculate the molality of solution.

2.27

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1.27

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3.27

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4.27

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Solution

The correct option is A $2.27$
Given,
Density= $1.017 {g m L}^{- 1}$
Molecular Mass= $60$
Molarity= $2.03 M$

$\to M o l a r i t y = 2.03 M$ means $2.03$ moles in $1 L$ of solution.

$n = \frac{G i v e n M a s s}{M o l a r M a s s}$

$2.03 = \frac{G i v e n M a s s}{60}$

$2.03 \times 60 = 121.8 g$

$121.8 g$ of acetic acid in water.

$\to D e n s i t y = 1.017 {g m L}^{- 1} 1 m L o f s o l u t i o n ⟶ 1.017 o f s o l u t i o n$

$1000 m L o f s o l u t i o n ⟶ 1017 g o f s o l u t i o n$

Mass of solution = Mass of solvent + Mass of solute

$1017 g$ = Mass of solvent + $121.8 g$

$1017 - 121.8 g$ = Mass of solvent

$895.2 g$ = Mass of solvent

$M o l a l i t y = \frac{n o . o f m o l e s}{M a s s o f s o l v e n t (k g)} = \frac{2.03}{895.2 \times 10^{- 3} k g} = 2.267 m$

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Density of 2.03M aqueous solution of acetic acid is 1.017gmL−1. Molecular mass of acetic acid is 60. Calculate the molality of solution.

Density of $2.03 M$ aqueous solution of acetic acid is $1.017 g {m L}^{- 1}$ . Molecular mass of acetic acid is $60$ . Calculate the molality of solution.