Question

Molarity of a dibasic acid is x(M),what is the normality of that acid?

Answer 1

Answer:

2x

Explanation:

normality = M×n- factor

n factor of dibasic acid is 2 as it can donate 2 H+

Answer 2

Answer:

The normality of that acid is 2x.

Explanation:

Molarity is the number of mole of the solute per unit volume of the solution.

$\text { Molarity }=\frac{\text { Moles of solute in solution }}{\text { Volume of solution }}$

But,

$\text { Mole of a solute }=\frac{\text { Mass of the solute }}{\text { Molecular weight }}$

Thus, Molarity, M is given by:

\text { Mole of a solute }=\frac{\text { Mass of the solute }}{\text { Molecular weight }}

$\text { Mole of a solute }=\frac{\text { Mass of the solute }}{\text { Molecular weight }}$

Where, m = mass of solute

V = Volume of solution.

Normality is the gram equivalent of the solute per unit volume of the solution.

$\mathrm{N}=\frac{\text {gram equivalent of solute}}{\text {Volume of solution}}$

$\mathrm{N}=\frac{g}{v}$

Where, g = gram equivalent of the solute

N = Normality.

We know that

$g=\frac{m}{\text {equivalent weight}}$

By definition,

$\text { Equivalent weight }=\frac{\text {Molar weight}}{n}$

Where, n = n-factor (change in valency/oxidation for ion/acidity/basicity)

Substituting the above in equation (2), we get

$\mathrm{N}=\frac{n m}{v+m o l a r \text {weight}}$

Combining equations (1) and (3), we get,

N = nM

Now, based on the question, here

M = x M

Since the given acid is dibasic, the amount of H+ it can release is 2. So,  

n = 2

Therefore, normality is given by  

$\mathrm{N}=2 * \mathrm{x}=2 \mathrm{x}$.