The ratio of equivalent weights of C_2H_5OH i | vedclass

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(C) In reaction $(i)$,$C_2H_5OH$ undergoes oxidation to $CH_3CHO$. The change in oxidation state of carbon is from $-2$ to $-1$. Since there are two c

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$1 : 2$

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(C) In reaction $(i)$,$C_2H_5OH$ undergoes oxidation to $CH_3CHO$. The change in oxidation state of carbon is from $-2$ to $-1$. Since there are two carbon atoms,the total change in oxidation number is $2 	imes ((-1) - (-2)) = 2$. Thus,the $n$-factor is $2$,and the equivalent weight is $E_1 = \frac{M}{2}$.In reaction $(ii)$,$C_2H_5OH$ acts as an acid reacting with $Na$ to form $C_2H_5ONa$. The number of replaceable $H^+$ ions is $1$. Thus,the $n$-factor is $1$,and the equivalent weight is $E_2 = \frac{M}{1}$.The ratio of equivalent weights $E_1 : E_2 = \frac{M}{2} : \frac{M}{1} = 1 : 2$.

Isotope	$\%$ Natural Abundance	Molar Mass $(g/mol)$
$^{35}Cl$	$75.77$	$34.9689$
$^{37}Cl$	$24.23$	$36.9659$

The ratio of equivalent weights of $C_2H_5OH$ in the following reactions is
$(i)$ $C_2H_5OH \longrightarrow CH_3CHO$
$(ii)$ $C_2H_5OH \longrightarrow C_2H_5ONa$

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(in $u$)

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The ratio of equivalent weights of $C_2H_5OH$ in the following reactions is$(i)$ $C_2H_5OH \longrightarrow CH_3CHO$$(ii)$ $C_2H_5OH \longrightarrow C_2H_5ONa$