Question

Two bottles of A and B contains 1M and 1m aqueous solution ($\text{d =1 g m}{{\text{L}}^{\text{-1}}}$) of sulphuric acid respectively. Which of the following statements is true about A and B?
(A) A is more concentrated than B
(B) B is more concentrated than A
(C) Concentration of A. Conc. of B
(D) It is not possible to compare the concentration

Answer 1

Hint: The molarity and molality are the measures of the concentration of the solution. The molarity is the number of moles per litre of solution and molality is the number of moles of solute in 1 kilogram of solvent. When we convert the 1 molal solution in terms of molarity we should use the density relation to convert the mass of solvent and solute to the volume. The molal is a less concentrated solution than a molar solution.

Complete step by step solution:
We have two bottles A and B containing the aqueous solution of sulphuric acid.
Bottle A contains the 1 molar solution. We know that molarity is defined as the number of moles (n) of solute dissolved per litres of the solution. It is denoted by the M. The molarity can be mathematically expressed as follows,
$\text{Molarity (M) = }\dfrac{\text{No}\text{. of moles}}{\text{Volume of solution in liters}}\text{ = }\dfrac{\text{n}}{\text{V}}\text{ = }\dfrac{\text{Mass}}{\text{Molar mass}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{Volume in Liters (L)}}$
Since the number of moles is defined as the ratio of the mass of the solute to the molecular weight of the solute.$\text{no}\text{. of moles = }\dfrac{\text{Mass of solute}}{\text{Molar mass of solute}}$
The molarity of the solution has units of M or$\text{mol }{{\text{L}}^{\text{-1}}}$.
The molecular weight of sulphuric acid \[{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\] is,
\[\begin{align}
& \text{Mol}\text{.wt}\text{.}{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}=\text{ 2(1) + 32 + 4(16)} \\
& \text{ }=\text{ 2(1) + 32 + 4(16)} \\
& \text{ = 98 g mo}{{\text{l}}^{\text{-1}}} \\
\end{align}\]
The 1 molar solution of \[{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\] means that the 1 mole of \[{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\] is in the 1 litre or $1000\text{ mL}$ of solution.
i.e.
$1\text{ M = }\dfrac{98\text{ g}}{98\text{ g mo}{{\text{l}}^{\text{-1}}}}\text{ }\times \text{ }\dfrac{1}{1000\text{ mL}}$
Let us take a bottle B. It contains a 1 molal solution. The molality is defined as the amount of the substance of the solute dissolved in the 1 kilogram of the solvent. It is denoted by the’m’. The molality is stated as the follows,
$\text{Molarity (M) = }\dfrac{\text{No}\text{. of moles}}{\text{Solvent in kilograms}}\text{ = }\dfrac{\text{Mass}}{\text{Molar mass}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{Solvent in kilograms (Kg)}}$
In 1 molal solution, the 1 mole of \[{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\] is in the $1000\text{ g}$ of the solvent. Since we are given that the density of the solvent is$1\text{ g m}{{\text{L}}^{\text{-1}}}$.
Let us convert the 1 molal solution to the molar solution.
Now we know that 1 molal is equal to the 1 moles in the 1 kilogram of solvent.
$1\text{ molal = 1 moles in 1 kg solvent}$
The total mass of the solution is equal to the mass of solvent and mass of solute \[{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\]
i.e.$\text{ Total mass of solution = 1000 g of solvent + }{{\text{M}}_{\text{solute}}}$
Since we are given that the density of the solution is $1\text{ g m}{{\text{L}}^{\text{-1}}}$.thus, the mass is equal to the volume of the solution.
Therefore, we can write,
\[\begin{align}
& \text{Volume of solution = 1000 + }{{\text{M}}_{\text{solute}}}\text{ mL} \\
& \text{ = 1000 + 98 }\because {{\text{M}}_{\text{solute}}}\text{=}{{\text{M}}_{{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}}}\text{= 98} \\
& \text{ = 1098 mL} \\
\end{align}\]
Thus, the molarity of 1 molal solution is,
$\begin{align}
& \text{Molarity of 1 molal solution = }\dfrac{\text{1}}{\dfrac{\text{1000+}{{\text{M}}_{\text{Solute}}}}{\text{1000}}}\text{ } \\
& \text{ = }\dfrac{\text{1}}{\text{1+}\dfrac{{{\text{M}}_{\text{Solute}}}}{\text{1000}}} \\
& \text{ = }\dfrac{\text{1}}{\text{1+}\dfrac{\text{98}}{\text{1000}}} \\
& \text{ = }\dfrac{\text{1}}{\text{1}\text{.098}} \\
& \text{Molarity of 1 molal solution = 0}\text{.910 M } \\
\end{align}$
Here, we get that the molarity of 1 molal solution is$\text{0}\text{.910 M }$.
The concentration of molarity is greater than that of the molality.
$\text{Molarity (M) }\rangle \text{ Molality (m)}$
Therefore, A is more concentrated than B.

Hence, (A) is the correct option.

Note: In molality, the mass written in the denominator is only of the solvent. Do not confuse it with the mass of solute and solvent together. Molality is considered the best in terms of experiencing the concentration of the solution. This is because molarity depends on the volume of the solution and thus changes with the temperature. But molality does not change with temperature as it depends on the mass of the solvent.