Question

Two bottles of A and B contains $1{\text{ M}}$ and $1{\text{ m}}$ aqueous solution $\left( {{\text{d}} = {\text{1g/mL}}} \right)$ of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}$ solution respectively.
A) A is more concentrated than B
B) B is more concentrated than A
C) Concentration of A = concentration of B
D) It is not possible to compare the concentration.

Answer 1

We are given that the two bottles of A and B contains $1{\text{ M}}$ and $1{\text{ m}}$ aqueous solution of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}$ solution. $1{\text{ M}}$ suggests that the concentration is one molar and $1{\text{ m}}$ suggests that the concentration is one molal. We know that molarity is the number of moles of solute per litre of solvent. To solve this we can convert the molal concentration to molar and then compare the both.

Complete step-by-step solution :
We are given that the two bottles of A and B contains $1{\text{ M}}$ and $1{\text{ m}}$ aqueous solution of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}$.
Consider bottle A which contains $1{\text{ M}}$ aqueous solution of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}$.
We know that molarity is the number of moles of solute dissolved per litre of solvent. The mathematical expression for molarity is as follows:
$\Rightarrow {\text{Molarity}}\left( {\text{M}} \right) = \dfrac{{{\text{Number of moles of solute}}\left( {{\text{mol}}} \right)}}{{{\text{Volume of solution}}\left( {\text{L}} \right)}}$
We know that the number of moles is the ratio of mass to the molar mass. Thus,
$\Rightarrow {\text{Molarity}}\left( {\text{M}} \right) = \dfrac{{{\text{Mass}}\left( {\text{g}} \right)}}{{{\text{Molar mass}}\left( {{\text{g/mol}}} \right)}} \times \dfrac{{\text{1}}}{{{\text{Volume of solution}}\left( {\text{L}} \right)}}$
The units of molarity are ${\text{mol/litre}}$ or ${\text{M}}$.

Now we will calculate the molar mass of acetic acid, ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}$ as follows:
Molar mass of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}$ $= \left( {2 \times {\text{C}}} \right) + \left( {2 \times {\text{O}}} \right) + \left( {4 \times {\text{H}}} \right)$
Molar mass of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}$ $= \left( {2 \times 12} \right) + \left( {2 \times 16} \right) + \left( {4 \times 1} \right)$
Molar mass of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}$ $= \left( {24} \right) + \left( {32} \right) + \left( 4 \right)$
Molar mass of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}$ $= 60{\text{ g/mol}}$
Thus, the molar mass of acetic acid, ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}$ is $60{\text{ g/mol}}$.
One molar i.e. $1{\text{ M}}$ solution of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}$ means that one mole of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}$ are dissolved in one litre of solution. One mole of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}$ contains $60{\text{ g}}$ of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}$. Thus,
$\Rightarrow {\text{Molarity}}\left( {\text{M}} \right) = \dfrac{{60{\text{ g}}}}{{60{\text{ g/mol}}}} \times \dfrac{{\text{1}}}{{{\text{1 L}}}} = 1{\text{ M}}$
Consider bottle B which contains $1{\text{ m}}$ aqueous solution of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}$.
We know that molality is the number of moles of solute dissolved per kilogram of solvent. The mathematical expression for molality is as follows:
$\Rightarrow {\text{Molality}}\left( {\text{m}} \right) = \dfrac{{{\text{Number of moles of solute}}\left( {{\text{mol}}} \right)}}{{{\text{Weight of solvent}}\left( {{\text{kg}}} \right)}}$
We know that the number of moles is the ratio of mass to the molar mass. Thus,
$\Rightarrow {\text{Molality}}\left( {\text{m}} \right) = \dfrac{{{\text{Mass}}\left( {\text{g}} \right)}}{{{\text{Molar mass}}\left( {{\text{g/mol}}} \right)}} \times \dfrac{{\text{1}}}{{{\text{Weight of solvent}}\left( {{\text{kg}}} \right)}}$
The units of molality are ${\text{mol/kg}}$ or ${\text{m}}$.
Let us convert one molar solution to one molal solution as follows:
In $1{\text{ m}}$ solution, one mole of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}$ is dissolved in $1000{\text{ g}}$ of solvent. Thus, the total mass of the solution is the mass of solute and the mass of solvent. Thus,
Mass of solution $= \left( {1000 + 60} \right){\text{g}}$
Mass of solution $= 1060{\text{ g}}$
We are given that the density of the solution is ${\text{1g/mL}}$. Thus, the mass of the solution is equal to the volume of the solution. Thus,
Volume of solution $= 1060{\text{ mL}}$
Thus, the molarity of $1{\text{ m}}$ solution is,
Molarity of $1{\text{ m}}$ solution $= \dfrac{1}{{\dfrac{{1060}}{{1000}}}}$
Molarity of $1{\text{ m}}$ solution $= \dfrac{1}{{1.060}}$
Molarity of $1{\text{ m}}$ solution $= 0.9433{\text{ M}}$
Thus, the molarity of $1{\text{ m}}$ solution is $0.9433{\text{ M}}$.
Thus, bottle A contains $1{\text{ M}}$ solution of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}$ and bottle B contains $0.9433{\text{ M}}$ solution of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}$
Thus, we can say that A is more concentrated than B.

Thus, the correct answer is option (A) A is more concentrated than B.

Note: In molality, we use the mass of the solvent in the denominator and not the mass of solution. The molarity of a solution depends on the volume of the solution and the volume changes with temperature and thus, molarity changes with temperature. Molality depends on the mass of the solvent and thus, molality does not change with temperature.