Question

Two samples of sulfuric acid are labeled as 1 M ${{H}_{2}}S{{O}_{4}}$ and 1 m ${{H}_{2}}S{{O}_{4}}$. The density of the solution is 1.15 g /cc, then which among the following is more concentrated?
(a)- 1 M ${{H}_{2}}S{{O}_{4}}$
(b)- 1 m ${{H}_{2}}S{{O}_{4}}$
(c)- Both are equal
(d)- Cannot be predicted

Answer 1

Hint:1 M ${{H}_{2}}S{{O}_{4}}$ means the molarity of the solution (1 molar solution of sulfuric acid) and 1 m ${{H}_{2}}S{{O}_{4}}$ means the molality of the solution (1 molal solution of sulfuric acid). The molarity is calculated by the volume of the solution and the molality of the solution is calculated by the weight of the solvent.

Complete step-by-step answer:We are given two solutions, 1 M ${{H}_{2}}S{{O}_{4}}$ and 1 m ${{H}_{2}}S{{O}_{4}}$.
Let us take the sample 1 = 1 M
Let us take the sample 2 = 1 m
Given density is 1.15 g /cc.
Now, let us assume that the volume of the sample is 1 liter = 1000 ml.
We know the molecular mass of the sulfuric acid (${{H}_{2}}S{{O}_{4}}$) = 98 g /mol.
The mass of the substance is equal to the product of the volume and density of the substance.
The mass of the sample will be = 1.15 g /cc x 1000 ml = 1150 g.
As we know that the molarity of the solution is equal to the ratio of moles of the solute to the volume of the solution. Then the number of moles of sulfuric acid will be = molarity x volume.
$Moles=1M\text{ x 1L = 1 mole}$
In sample 1, the mass of the solute will be = no. of moles x molar mass of ${{H}_{2}}S{{O}_{4}}$.
$\text{Mass of }{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}=\text{ 1 x 98 = 98 g}$
Since we know the mass of solution and mass of solute, so we can calculate the mass of solvent. Mass of solvent = 1150 – 98 = 1052 g = 1.052 kg.
Now, we can calculate the molality of sample 1:
$Molality=\dfrac{1}{1.052}=0.951\text{ m}$
So, the molality of sample 1 is less than the molality of sample 2, therefore, sample 2 has more concentration.

Hence, the correct answer is an option (b).

Note:The number of moles is calculated by taking the ratio of the given mass to the molecular mass of the compound. The denominator for molality is the mass of solvent and for molarity is the volume of the solution.