Question

For $2 \times {10^{24}}$ molecules of ${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$ is present in 500mL of solution. What is the molarity of the solution?

Answer 1

Hint: In the International System of Units, the mole is the unit of measurement for the volume of matter. It is known as a set of exactly \[6.02214076 \times {10^{23}}\] particles, which may be atoms, molecules, ions, or electrons. The mole is basically a particle count.

Complete answer:
The particles being counted are usually chemically similar structures that are uniquely distinct. A solution, for example, could have a certain amount of dissolved molecules that are more or less independent of one another. The constituent particles of a solid, on the other hand, are set and bound in a lattice structure, but they can be separated without losing their chemical identity.
Molar concentration (also known as molarity, quantity concentration, or material concentration) is a measurement of a chemical species' concentration in a solution in terms of the amount of substance per unit volume of solution. The number of moles per litre, abbreviated as \[mol.d{m^{ - 3}}\] in SI units, is the most widely used metric for molarity in chemistry. A solvent with a concentration of 1 mol/L is known as a molar solution, or 1 M. Small caps or italicised M are often used in publications and textbooks to prevent confusion with the SI prefix mega, which has the same abbreviation.
${\text{Molarity = }}\dfrac{{{\text{Number of moles}}}}{{{\text{Volume in litres}}}}$
Now we need to find the number of moles in $2 \times {10^{24}}$molecules of ${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$
${\text{Number of moles = }}\dfrac{{{\text{Given Number of Molecules}}}}{{{N_A}}}$
${\text{Number of moles = }}\dfrac{{2 \times {{10}^{24}}}}{{6.023 \times {{10}^{23}}}}$
${\text{Number of moles = }}\dfrac{{20 \times {{10}^{23}}}}{{6.023 \times {{10}^{23}}}} = 3.32moles$
Now to find the molarity,
Volume = 500mL = 0.5 L
Number of moles = 3.32 moles
Substituting them we get,
${\text{Molarity = }}\dfrac{{{\text{Number of moles}}}}{{{\text{Volume in litres}}}} = \dfrac{{3.32}}{{0.5}}$
${\text{Molarity }} = \dfrac{{3.32}}{{0.5}} = 6.67$M

Note:
The use of molar concentration in thermodynamics is sometimes inconvenient since the volume of most solutions varies significantly with temperature due to thermal expansion. This problem is usually solved by using temperature adjustment factors or a temperature-independent concentration measure such as molality. The reciprocal quantity denotes the dilution (volume) that can be found in Ostwald's dilution law.