Moles of N in N, N - dimethylaminopentane = $$\left( {{{57.5} \over {115}}} \right)$$ = 0.5 mol

$$ \Rightarrow {C_7}{H_{17}}N + {{45} \over 2}CuO \to 7C{O_2} + {{17} \over 2}{H_2}O + {1 \over 2}{N_2} + {{45} \over 2}Cu$$

$${{{n_{CuO}}\,reacted} \over {\left( {{{45} \over 2}} \right)}} = {{{n_{{C_7}{H_{17}}N}}\,reacted} \over 1}$$

$$\Rightarrow$$ n_CuO reacted = $$\left( {{{45} \over 2}} \right) \times 0.5 = 11.25$$

Aqueous tension at 287 K = 14 mm of Hg.

Hence actual pressure = (758 – 14)
= 744 mm of Hg.

Moles of $${N_2} = {{\left( {758 - 14} \right)} \over {760}} \times {{30 \times {{10}^{ - 3}}} \over {0.0821 \times 287}}$$

$$ = 1.246 \times {10^{ - 3}}$$ mol

mass of $${N_2} = 1.246 \times {10^{ - 3}} \times 28$$

mass % of N₂ $$ = {{mass\,of\,'N'} \over {total\,mass}} \times 100$$

$$ = {{1.246 \times 28 \times {{10}^{ - 3}}} \over {0.184}} \times 100$$

$$ = {{124.6 \times 28} \over {0.184}}\% = 18.96\% $$

$$ \simeq 19\% $$

$${R_f} = {{Dis\tan ce\,travelled\,by\,compound} \over {Dis\tan ce\,travelled\,by\,solvent}}$$

On chromatogram distance travelled by compound is = 2 cm

Distance travelled by solvent = 5 cm

So, $${R_f} = {2 \over 5} = $$ 4 $$\times$$ 10^$$-$$1 = 0.4

Given, weight = 18.6 g

Here, 1 mole of aniline gives 1 mole of acetanilide

$$\therefore$$ mole of aniline = mole of acetanilide

$$\Rightarrow$$ $${{1.86} \over {93}} = {{{W_{Acetanilide}}} \over {135}}$$

$${W_{Ace\tan ilide}} = {{1.86 \times 135} \over {93}}g = 2.70g$$

But efficiency of reaction is 90% only.

Hence, mass of acetanilide produced $$ = 2.70 \times {{90} \over {100}}g = 2.43g = 243 \times {10^2}g$$

x = 243