Question

100mL of a mixture of NaOH and Na2SO4 is neutralized by 10mL of 0.5M H2SO4. Hence NaOH in 100mL solution is

Answer 1

NaOH in thus case undergoes neutralisation.

The reaction equation for neutralization is :

2NaOH + H₂SO₄ —> Na₂SO₄ + 2H₂O

Mole ratio is :

2 : 1

Moles of sulphuric acid:

(10 / 1000 ) × 0.5 = 0.005 moles

Moles of sodium hydroxide :

0.005 × 2 = 0.01 moles.

Molarity of NaOH :

(1000/100) × 0.01 = 0.1 M

Answer 2

Answer: 4 gm

Explanation:NaOH in thus case undergoes neutralisation.

The reaction equation for neutralization is :

2NaOH + H₂SO₄ —> Na₂SO₄ + 2H₂O

Mole ratio is :

2 : 1

Moles of sulphuric acid:

(10 / 1000 ) × 0.5 = 0.005 moles

Moles of sodium hydroxide :

0.005 × 2 = 0.01 moles.

Molarity of NaOH :

(1000/100) × 0.01 = 0.1 M

0.1*40=4