Chemistry - Concentration Terms Question with Solution | TestHub

ChemistryConcentration TermsLabeling of OleumMedium2 minQB

ChemistryMediumsingle choice

Passage / Comprehension

A sample of fuming sulphuric acid containing $H_{2} SO_{4}, SO_{3}$ and $SO_{2}$ weighing 1450 g is found to require 18 L of 2.00 M alkali $(NaOH)$ for neutralisation. A separate sample shows the presence of 5 mole $SO_{2}$ .

Find the percentage of $H_{2} SO_{4}$ in sample :

Options:

Answer:

Solution:

Let moles of H₂SO₄ = $x$ and SO₃ = $y$ .

NaOH used = $18 \times 2 = 36$ mol.

Neutralization: H₂SO₄ consumes 2 mol NaOH, and SO₃ (converted to H₂SO₄) consumes 2 mol NaOH, so:

$2 (x + y) = 36 \Rightarrow x + y = 18$

Given SO₂ = 5 mol, mass:

$5 \times 64 = 320 g$

Total mass:

$98 x + 80 y + 320 = 1450$

Using $x + y = 18$ :

$98 x + 80 (18 - x) = 1130 \Rightarrow x = 5$

Mass of H₂SO₄:

$5 \times 98 = 490 g$

$% H_{2} S O_{4} = \frac{490}{1450} \times 100 = 33.79%$

Stream:JEESubject:ChemistryTopic:Concentration TermsSubtopic:Labeling of Oleum

⏱ 2mℹ️ Source: QB

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