23. 4 g of solid NaOH was dissolved in water and the solution was made up to 1000 ml . The whole of this will neutralized completely

Question

 4 g of solid NaOH was dissolved in water and the solution was made up to 1000 ml . The whole of this will neutralized completely .

1. 100ml of 1 M H₂SO₄
2. 20 ml of 2.5M H₂SO₄
3. 20 ml of 1.5 M H₂SO₄
4. 30 ml of 5 M H₂SO₄.​

Answer 1

Correct option is (B) 20 ml of 2.5M H₂SO₄

Mass of solid NaOH = 4g

Molecular mass of NaOH = 40g/mol

So, number of mole of NaOH = 4/40 =0.1

Molarity = number of mole of solute/volume of solution in litre

= 0.1 × 1000/1000

= 0.1M

Then, normality = n-factor × molarity

= 1 × 0.1 

= 0.1N

For completely neutralization:

Equivalents of acid = Equivalents of base

or, n- factor of acid × mole of acid = 1 × 0.1 = 0.1

[for H₂SO₄, n-factor = 2]

or, mole of acid = 0.1/2 = 0.05

Now check options,

option (A) ⇒ mole of acid = 1 × 100/1000 = 0.1

option (B) ⇒ mole of acid = 2.5 × 20/1000 = 50/1000 = 0.05 

Hence, option (B) is satisfied the above condition.