One mole of phenol is treated with dilute HNO | vedclass

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(D) Phenol $(C_6H_5OH)$ reacts with dilute $HNO_3$ to form a mixture of $o$-nitrophenol and $p$-nitrophenol.$o$-Nitrophenol is steam volatile due to i

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$45$

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(D) Phenol $(C_6H_5OH)$ reacts with dilute $HNO_3$ to form a mixture of $o$-nitrophenol and $p$-nitrophenol.$o$-Nitrophenol is steam volatile due to intramolecular hydrogen bonding, while $p$-nitrophenol is not due to intermolecular hydrogen bonding.Thus, the steam volatile compound $(X)$ is $o$-nitrophenol $(C_6H_5NO_3)$.Molar mass of phenol $(C_6H_6O)$ = $(6 	imes 12) + (6 	imes 1) + 16 = 94 \ g \ mol^{-1}$.Percentage of oxygen in phenol = $(16 / 94) 	imes 100 \approx 17.021 \ \%$.Molar mass of $o$-nitrophenol $(C_6H_5NO_3)$ = $(6 	imes 12) + (5 	imes 1) + 14 + (3 	imes 16) = 72 + 5 + 14 + 48 = 139 \ g \ mol^{-1}$.Percentage of oxygen in $o$-nitrophenol = $(48 / 139) 	imes 100 \approx 34.532 \ \%$.Increase in percentage of oxygen = $34.532 - 17.021 = 17.511 \ \%$.Expressing as $175.11 	imes 10^{-1} \% \approx 175 	imes 10^{-1} \%$. Given the options provided, the closest integer value is $175$ (Note: The provided options in the prompt seem to be placeholders; based on standard calculation, the result is $175$).

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