Question

How are \[0.50\] mol $N{a_2}C{O_3}$ and $0.50M$ $N{a_2}C{O_3}$ different from each other?

Answer 1

Hint: We can solve this problem with the help of the definition of moles and molarity. Mole can be defined as that one mole is the amount of substance that contains as many entities or particles as there are present in $12g$ of ${C_{12}}$ atom which is exactly equal to $6.02 \times {10^{23}}$ particles, where $6.02 \times {10^{23}}$ is the Avogadro number and molarity is defined as the number of moles of a solute dissolved in one litre of solution. Molarity is the measure of concentration of a chemical species.

Complete step-by-step answer:
> So first of all we will calculate molar mass of sodium carbonate .Molar mass sodium carbonate $ = 2 \times $ atomic mass of sodium $\left( {Na} \right)$ $ + $ atomic mass of carbon $\left( C \right) + 3 \times $ atomic mass of oxygen $(O)$ $ = 2 \times 23 + 12 + 3 \times 16 = 106$ gram per mole. Thus this shows that one mole of sodium carbonate is equal to $106$ gram.
> So \[0.50\] mol $N{a_2}C{O_3}$ $ = 0.5mole \times 106g/mole = 53$ gram hence \[0.50\] mol $N{a_2}C{O_3}$ means half of molar mass of sodium carbonate that is $53$ gram and it also indicates $3.01 \times {10^{23}}$ molecules.
> By the definition of molarity we can say that $0.50M$ $N{a_2}C{O_3}$ means that \[0.50\] mol of sodium carbonate in one litre of solution.
> Hence we can say that \[0.50\] mol $N{a_2}C{O_3}$ and $0.50M$ $N{a_2}C{O_3}$are different as are \[0.50\] mol $N{a_2}C{O_3}$ states that it has $3.01 \times {10^{23}}$ molecules and $0.50M$ $N{a_2}C{O_3}$ states that \[0.50\] mol or $53$ gram of sodium carbonate is present in one litre of water.

Note: In the mole concept we have studied that mole , molarity and molality etc. all these terms are different but they share some relationship with each other. In this problem \[0.50\] mol $N{a_2}C{O_3}$ and $0.50M$ $N{a_2}C{O_3}$ are different. We know that molarity can not be defined without defining moles so they all are related.