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The molarity of 0.2 N $N{{a}_{2}}C{{O}_{3}}$ solution will be:a.) 0.05 Mb.) 0.2 Mc.) 0.1 Md.) 0.4 M

Last updated date: 07th Aug 2024
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Hint: If we know the normality of a solution then we can convert it into normality by dividing normality by the number of hydrogen ions in acid or base.

Molarity is defined as the number of moles of solute present in one litre of solution. Normality is defined as the number of molecules equivalent to solute in litre of the solution.
Normality is calculated using the formula
Normality = $\dfrac{\text{gram equivalent weight}}{\text{litre of solution}}$
And formulae for molarity is

Molarity= $\dfrac{\text{moles of solute}}{\text{litre of solution}}$

To convert normality to molarity or molarity to normality we have to find the n factor which is the number of hydrogen ions in acid or base.
But in this question sodium carbonate is a salt and to find the n factor of a salt we have to calculate the number of moles of electrons exchanged either it is loosed or gained by the one mole of the given salt which is sodium carbonate.
There is an exchange of two electrons in sodium carbonate hence the n factor of sodium carbonate is 2.
Given in the question
Normality of sodium carbonate = 0.2 N
The formulae that defines the relation between normality and molarity is normality is equal to the product of n factor and molarity.
But here normality is given to molarity will be equal to the ration of normality to n factor
Molarity = $\dfrac{\text{normality}}{\text{n factor}}=\dfrac{0.2}{2}$
Molarity = 0.1 M
So, the correct answer is “Option C”.

Note: Pay attention while calculating n factor, there are different ways to calculate n factor for acid, base or a salt or chemical reaction. The n factor for a redox reaction is equal to the number of moles of lost or gained electrons per molecule.