Question

Density of a $2.05M$ solution of acetic acid in water is $1.02g/mL$ . The molality of the solution is:
A. $1.14molk{g^{ - 1}}$
B. $3.28molk{g^{ - 1}}$
C. $2.28molk{g^{ - 1}}$
D. $0.44molk{g^{ - 1}}$

Answer 1

Molarity is defined as the moles of solute dissolved per litre of the solution and molality is defined as the moles of solute per kg of the solution.

Complete step by step answer:
Molarity is the measure of concentration and is expressed as the ratio of number of moles of solute and volume of solution in liters. It is denoted as $M$.
$M = \dfrac{n}{V}$
where $n$ is the number of moles of solute and $V$ is the volume of the solution in litres.
Molality is another way of measurement of concentration and is expressed as the number of moles of solute per kg of the solvent. It is denoted as $m$.
$m = \dfrac{n}{W}$
where $n$ is the number of moles of solute and $W$ is the weight of the solvent in kilogram.
The difference between molarity and molality is liters and kg. Thus both the molarity and molality are related to density. Density is the ratio of mass upon volume and is expressed as
$d = \dfrac{w}{V}$
Where w is the mass of the substance and V is the volume.
Given, density is $1.02g/mL$ , thus the weight or mass of the solution = $d \times V$
=$1.02 \times 1000 = 1020g$.
The molar mass of acetic acid ($C{H_3}COOH$ ) = $2 \times C + 4 \times H + 2 \times O$
$24 + 4 + 32 = 60g$.
The weight of acetic acid in solution = $2.05 \times 60 = 123g$.
The weight of water = $1020 - 123 = 897g$.
Thus molality = $\dfrac{{2.05}}{{897}} \times 1000 = 2.28$.
Hence option C is the correct answer, i.e. molality of the solution is $2.28molk{g^{ - 1}}$.

Note:
Molality is a favoured measurement for the concentration of a solution over molarity. Molarity of a solution varies with volume as the volume is dependent on temperature. This causes more errors in measuring molarity.