Question

How$0.50\mathrm{mol}{\mathrm{Na}}_2{\mathrm{CO}}_3$is different from$0.50\mathrm{M}{\mathrm{Na}}_2{\mathrm{CO}}_3$?

Answer 1

Molarity(M):-

• Molarity is the number of moles of solutes dissolved in one liter of solution.
• It is denoted by $\mathrm{M}$ and its unit is $\mathrm{mol}{\mathrm{L}}^{-1}$.
• The molarity is calculated by the following formula,

$\mathrm{Molarity}=\frac{\mathrm{Number}\mathrm{of}\mathrm{moles}}{\mathrm{Volume}\mathrm{of}\mathrm{solution}\mathrm{in}\mathrm{liter}}$

Mole:-

• Moles are a number of particles, atoms, and molecules present in a given amount of substance as the number of atoms present in 12g of Carbon - 12 isotope.
• It is denoted by $\mathrm{mol}$ and is a unitless quantity.
• It is calculated by the following formula,

$\mathrm{Number}\mathrm{of}\mathrm{moles}=\frac{\mathrm{Given}\mathrm{mass}\mathrm{of}\mathrm{substance}}{\mathrm{Atomic}\mathrm{mass}\mathrm{of}\mathrm{substance}}$

Difference between $0.50\mathrm{mol}{\mathrm{Na}}_2{\mathrm{CO}}_3(\mathrm{Sodium}\mathrm{carbonate})$ and $0.50\mathrm{M}{\mathrm{Na}}_2{\mathrm{CO}}_3$:

1. $0.50\mathrm{moles}$ of ${\mathrm{Na}}_2{\mathrm{CO}}_3$ means, $0.50$ moles of ${\mathrm{Na}}_2{\mathrm{CO}}_3$ are present in the reaction mixture.
2. $0.50\mathrm{M}{\mathrm{Na}}_2{\mathrm{CO}}_3$ means $0.50\mathrm{moles}$ of Sodium carbonate are present in the one liter of solution.

Therefore, the difference between $0.50\mathrm{mol}{\mathrm{Na}}_2{\mathrm{CO}}_3$ and $0.50\mathrm{M}{\mathrm{Na}}_2{\mathrm{CO}}_3$ is the amount of substance present in a given reaction mixture and in a solution of one liter.