Skip to main content
Back to PRACTICAL CHEMISTRY

JEE CHEMISTRY MCQ: PRACTICAL CHEMISTRY

1 Medium
1 of 1
Analyzing +4 | -1
A substance on treatment with dil. H2SO4\mathrm{H}_2 \mathrm{SO}_4 liberates a colourless gas which produces (i) turbidity with baryta water and (ii) turns acidified dichromate solution green. The reaction indicates the presence of
View answer and explanation
Correct option: C
SO32−\mathrm{SO}_3^{2-}
The correct answer is C (SO32−\mathrm{SO}_3^{2-}) because the colorless gas that produces turbidity with baryta water (Ba(OH)2\mathrm{Ba(OH)_2}) and turns acidified dichromate green is sulfur dioxide (SO2\mathrm{SO_2}), which is released from sulfite ions (SO32−\mathrm{SO}_3^{2-}).

More Questions from PRACTICAL CHEMISTRY

Practice All Chapter MCQs →
Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.