In electrolytic cell a sample of naphthalene solution is subjected to electrolysis with 1 amp current of 96.5% efficiency for 100 sec. $H_{2} O$ is electrolysed at negative metal plate and naphthalene is electrolysed at positive metal plate to produce pthalic acid and $C O_{2}$ . Total pthalic acid produced at electrode is completely neutralized by using a ‘x’ g 10% by weight pure NaOH sample.
Information $\Rightarrow$
a) Napthalene

b) Pthalic acid
c) Consider naphthalene is soluble in water.
d) Molar volume at NTP, STP are 22.4 lit, 22.7 lit.
Which of the following is/are correct?

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Volume of $C O_{2}$ produced at anode at STP is 2.52 ml

Value of ‘x’ is 0.022 g

Mass of pthalic acid produced is 9.22 mg.

Volume of $H_{2}$ produced at cathode at NTP is 11.2 ml

answer is A, B, C.

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Detailed Solution

$\frac{1 \times 96.5 \times 100}{100 \times 96500} = 10^{- 3} F \begin{array}{l} {}^{m}H_{2} = 1 \times 10^{- 3} = 10^{- 3} g; {}^{n}H_{2} = \frac{1}{2} \times 10^{- 3} g \\ {}^{v}H_{2} = \frac{1}{2} \times 10^{- 3} \times 22, 400 = 11.2 m l \end{array}$

b) Question Image

$\begin{array}{l} 8 H_{2} O + C_{10} H_{8} \to C_{8} H_{6} O_{4} + 2 C O_{2} + 18 H^{+} + 18 e \\ 18 F \equiv 2 m o l e s C O_{2} \\ 10^{- 3} F \equiv n_{C O_{2}} \\ n_{C O_{2}} = \frac{10^{- 3} \times 2}{18}; V_{C O_{2}} = \frac{10^{- 3} \times 2}{158} \times 22, 700 \\ V_{C O_{2}} = 2.52 m l \end{array}$

c) $18 F \equiv 166 g$

$\begin{array}{l} 10^{- 3} F \equiv w \\ w = \frac{10^{- 3} \times 166}{18} \end{array}$

c) Question Image

$166 g \equiv 80 g N a O H$

$\begin{array}{l} 9.2 \times 10^{- 3} g \equiv ? \\ \Rightarrow \frac{9.2 \times 10^{- 3} \times 80}{166} = \frac{4.4 g}{1000} 10 = \frac{4.4}{W_{s a m p l e}} \times 100 \end{array}$