An organic compound "x" where molar ratio of C, O and H are equal, on treatment with 50% KOH under reflux followed by acidification produced "y". The most likely structure of "y" is :
[Molar mass of 'x' is 58 g mol$$^{-1}$$]

Step 1:

Let the molar ratios of the elements in compound $$x$$ be equal.
So its empirical formula is $$C_nH_nO_n$$.

Molar mass calculation:
Atomic masses: C = 12, H = 1, O = 16.
Molar mass of $$C_nH_nO_n = 12n + 1n + 16n = 29n$$.

Given molar mass of $$x$$ is 58 g mol$$^{-1}$$, 

Therefore,
$$29n = 58 \;\Longrightarrow\; n = 2$$.

Hence the molecular formula of $$x$$ is $$C_2H_2O_2$$.

Step 2:

Saturation check (index of hydrogen deficiency):
For two carbons the saturated formula would be $$C_2H_{2\times2+2}=C_2H_6$$.

Degrees of unsaturation $$=\dfrac{(2C+2-H)}{2}=\dfrac{(4+2-2)}{2}=2$$.

Two degrees of unsaturation can be satisfied by two carbonyl groups, giving the structure $$OHC{-}CHO$$ (glyoxal). Glyoxal has no $$\alpha$$-hydrogen, so it undergoes Cannizzaro reaction.

Reaction in 50 % KOH, reflux:

Two molecules of glyoxal disproportionate (Cannizzaro reaction): one carbonyl group is reduced to an alcohol and the other is oxidised to a carboxylate within the same carbon skeleton, giving potassium glycolate $$HO{-}CH_2{-}COOK$$.

Subsequent acidification converts the salt to its free acid:

$$HO{-}CH_2{-}COOK + HCl \longrightarrow HO{-}CH_2{-}COOH + KCl$$.

Thus product $$y$$ is glycolic acid, $$HO{-}CH_2{-}COOH$$.

Option C $$\longrightarrow\ $$ $$O = C(OH){-}CH_2{-}OH$$ (glycolic acid)