CBSE Class 11-science Answered

how to solve this ?

Asked by arijitbhkt | 12 Sep, 2018, 09:26: AM

Expert Answer

The reaction equations are as :

(NH₄)₂SO₄ + 2NaOH → 2NH₃ + Na₂SO₄ + 2H₂O ------ (1)

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O ------ (2)

Now for 5 ml 0.2 N H₂SO_4,

0.2 N H₂SO₄ = 0.1 M H₂SO₄

No. of moles of H₂SO₄ = molarity × Volume in litre

= 0.1 × 0.005

No. of moles of H₂SO₄ = 0.0005 mol

From reaction (2)

As 1 mole of H₂SO₄ requires 2 mole of NaOH

Therefore 0.0005 moles of H₂SO₄ will require 0.001 mole of NaOH

25 ml solution is taken for titration, which is 1/10 th of the original solution.

Therefore, no. of moles of NaOH remained after the reaction with is 0.001×10 = 0.01 mole

No. space of space moles space of space NaOH space initially space present space equals fraction numerator 0.2 cross times 100 over denominator 1000 end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 0.02 space moles Mole space of space NaOH space Reacted space with space left parenthesis NH subscript 4 right parenthesis subscript 2 SO subscript 4 space space equals space 0.02 minus 0.01 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 0.01 space mole

From reaction (1) 2 moles of NaOH reacts with 1 mole of (NH₄)₂SO₄

Therefore 0.01 mole of NaOH will react with 0.005 moles of (NH₄)₂SO₄

We have,

No. space of space moles space equals fraction numerator Weight over denominator No. space of space moles end fraction Weight space equals space No. space of space moles cross times No. space of space moles space space space space space space space space space space space equals 0.005 space cross times 132.14 space space space space space space space space space space space equals space 0.66 space gm percent sign space Purity space equals space fraction numerator 0.66 over denominator 0.7 end fraction cross times 100 space space space space space space space space space space space space space space equals space 94.28 space percent sign

Percentage purity of (NH₄)₂SO₄ is 94.28%

Answered by Varsha | 12 Sep, 2018, 02:03: PM

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