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Standard XII

Chemistry

Synthesis Using Azide & Gabriel Synthesis

What volume o...

Question

What volume of 1 M NaOH (in ml) will be required to react completely with 100g of oleum which is 109% labelled?

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Solution

109% oleum means that 9g of H₂O is required to convert all SO₃ present in 100g of it to H₂SO₄.

SO₃ + H₂O → H₂SO₄

1 mole (80g) of SO₃ reacts with 1 mole (18g) of H₂O

So 40g of SO₃ will react with 9g of H₂O.

According to the following equation, 80g of NaOH will react with 80g and 98g of SO₃ and H₂SO₄ respectively.

SO₃ + 2NaOH → Na₂SO₄ + H₂O

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

Wt. of SO₃ present in 100g of the oleum sample = 40g

Wt. of H₂SO₄ present in 100g of the oleum sample = (100 – 40)g = 60g

Wt. of NaOH required to completely neutralize 40g of SO₃= (80x40)/80 = 40g

Wt. of NaOH required to completely neutralize 60g of H₂SO₄ = (80x60)/98 = 48.98g

Total Wt. of NaOH required for complete neutralization = 40 + 48.98 = 88.98g

Volume of 1M NaOH solution containing 88.98g of NaOH = (1000x88.98)/40 = 2224.5mL

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