$$A \xrightarrow{(i) NaOH, (ii) H_3O^+} B \xrightarrow{(i) EtOH, (ii) H_2SO_4, \Delta} C$$
'A' shows positive Lassaign's test for N and its molar mass is 121. 'B' gives effervescence with aq. NaHCO$$_3$$. 'C' gives fruity smell. Identify A, B and C.

The sequence of reagents tells us step-by-step what functional group has to be present in each compound.

Step 1 : A → B by $$NaOH/H_3O^+$$
Alkaline hydrolysis followed by acidification converts an amide, nitrile or ester into a carboxylic acid. Because ‘A’ gives a positive Lassaigne’s test for nitrogen, the functional group in A must contain nitrogen. Among the common acid derivatives only amides and nitriles satisfy this.

Molar-mass check for A
Given molar mass of A = 121 g mol$$^{-1}$$. An aromatic amide of formula $$C_6H_5CONH_2$$ (benzamide) has
$$7\;C \;(7\times 12)=84,\; 7\;H=7,\; 1\;N=14,\; 1\;O=16$$
Total $$=84+7+14+16=121\;g\,mol^{-1}$$, exactly the required value. Hence $$A = C_6H_5CONH_2$$ (benzamide).

Step 2 : B gives effervescence with $$NaHCO_3$$
Only acids that are stronger than $$H_2CO_3$$ liberate $$CO_2$$ from $$NaHCO_3$$. The product B obtained after hydrolysis must therefore be a carboxylic acid: $$C_6H_5CONH_2 \xrightarrow{NaOH/H_3O^+} C_6H_5COOH$$ (benzoic acid). Thus $$B = C_6H_5COOH$$.

Step 3 : B → C by $$EtOH/H_2SO_4,\,\Delta$$
Concentrated $$H_2SO_4$$ and hot ethanol carry out Fischer esterification, converting a carboxylic acid into its ethyl ester, which typically has a pleasant fruity odour. $$C_6H_5COOH + HOCH_2CH_3 \xrightarrow{H_2SO_4,\;\Delta} C_6H_5COOCH_2CH_3 + H_2O$$ Hence $$C = C_6H_5COOCH_2CH_3$$ (ethyl benzoate), noted for its sweet fruity smell.

Therefore
A : benzamide $$\left(C_6H_5CONH_2\right)$$
B : benzoic acid $$\left(C_6H_5COOH\right)$$
C : ethyl benzoate $$\left(C_6H_5COOCH_2CH_3\right)$$.

Option A which is: A = benzamide, B = benzoic acid, C = ethyl benzoate.