Solution

Step 1. To calculate the $$\%$$ of $$N$$
Volume of $$0.05 \,M \,H_{2}SO_{4}$$ left unused $$= v \,ml$$
Volume of $$0.1 \,M \,NaOH$$ required for neutralization of excess acid $$= 25.0 \,ml$$

Applying molarity equation,
$$n_{a}M_{a}V_{a} (H_{2}SO_{4}) = n_{b} M_{b} V_{b} (NaOH)$$

i.e., $$2 \times 0.5 \times v = 1 \times 0.1 \times 25$$

or $$v = 25\,ml$$

$$\therefore$$ Volume of $$0.5 \,M \,H_{2}SO_{4}$$ used

$$= 50 - 25 = 25\,ml$$

$$\%N = \dfrac{1.4 \times 25 \times 0.05 \times 2}{0.303} = 11.55$$

Step 2. To determine the molecular formula of the compound

$$C : H : N : O = \dfrac{69.4}{12} : \dfrac{5.8}{1} : \dfrac{11.55}{14} : \dfrac{13.25}{16}$$

$$= 5.78 : 5.8 : 0.825 : 0.828$$

$$= 7 : 7 : 1 : 0$$

$$\therefore$$ E.F. of the compound $$= C_{7}H_{7}NO$$

E.F. wt. $$= 7 \times 12 + 7 \times 1 + 14 + 16 = 121$$

Mol. mass of the compound $$= 121$$

$$\therefore$$ M.F. of the compound $$= C_{7}H_{7}NO \times \dfrac{121}{121}$$

$$= C_{7}H_{7}NO$$

Step 3. To draw the possible structures of the compound:
Since the compound is aromatic, it must contain a $$C_{6}H_{5}$$ group. In other words, the compound may be written as $$C_{6}H_{5}CH_{2}NO$$ or $$C_{6}H_{5}CH = NOH.$$ Since oximes show geometrical isomerism therefore, possible structures of the compound are: