Question

Find the odd one out
(i)- Camphor, Ammonium Chloride, Naphthalene balls, Sugar
(ii)- Turmeric, Methyl Orange, Rose petals, Beetroot

Answer 1

Hint: In part (i), it is a sweet tasting carbohydrate. It is used in daily household work. It does not get converted into gaseous form easily.
In part (ii), it is used widely as an indicator in volumetric analysis. It is synthetically prepared.

Complete step by step solution:
Let us see what the answer is.
Camphor is a waxy flammable transparent organic compound. It is whitish in colour. It readily evaporates (gets converted into gaseous form) if left open. It is volatile in nature.
Ammonium Chloride is a white inorganic salt. It is widely used in the qualitative assessment of the salts. It readily decomposes into ammonia and HCl.
Naphthalene balls are an organic compound. It is a polycyclic aromatic hydrocarbon. It readily sublimes to vapour state. It is used as a fumigant.
Sugar is an organic carbonyl compound. It is a sweet tasting carbohydrate. It is quite strong in nature. It does not get directly converted into gaseous form.
Sugar is a non-volatile compound, so SUGAR is the odd one out.

Turmeric is a natural indicator. It gives red colour in basic medium and remains yellow coloured in acidic medium.
Rose petal pigment is used as an indicator. It gives dark red colour in acidic medium and reddish green colour in basic medium.
Beetroot contains anthocyanin which makes it an indicator. It gives red colour in acidic medium and blue colour in basic medium.
Methyl orange is a synthetically prepared compound. It gives red colour in acidic medium and yellow colour in basic medium.

Methyl orange is a synthetic indicator, so METHYL ORANGE is the odd one out.

Note: You may get confused in part (i) and mark Ammonium Chloride out of camphor, naphthalene balls and sugar as the answer considering the nature of compound i.e. organic or inorganic. But we have to see similarities between properties of compounds.
The pH range of Methyl Orange is 3.1 (red) to 4.4 (yellow).