Question

the normality of 500 ml of 0.2 M sulphuric acid is?​

Answer 1

Answer:

0.4

Explanation:

we know that normality=molarity x nfactor

so, 500=molarity x 0.2

molarity =0.4

Answer 2

Answer:

The normality of 500ml of sulphuric acid solution is equal to 0.4N.

Explanation:

Given: The molarity of solution of sulphuric acid $=0.2M$

The volume of solution $V= 500ml=0.5L$

Molarity of a solution can be calculated from the number of moles of the solute divided by volume of solution.  

$Molarity=\frac{moles\; of \; solute}{Volume \: of \: solution}$

Normality can be calculated from gram equivalents divided by volume of solution ( in L).

$Normality=\frac{(mass)*(n-factor)}{(Molar mass)\: *(volume)}$

$Normality=[\frac{(no.\: of \: moles)}{Volume\: of\: solution}] *(n-factor)$

Normality = Molarity × n-factor

Now, the n-factor for H₂SO₄ = number of H⁺ ions = 2

Therefore, Normality = 0.2× 2

Normality of  H₂SO₄ solution = 0.4N

Therefore, the normality of 500ml H₂SO₄ is equal to 0.4N.