The normality of 10 % (weight/volume) acetic | vedclass

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Question

(C) The formula for normality is $N = \frac{	ext{mass of solute in } g 	imes 1000}{	ext{Equivalent weight} 	imes 	ext{Volume of solution in } mL}

Vedclass · Accepted Answer

$1.7$

Vedclass · Answer

(C) The formula for normality is $N = \frac{	ext{mass of solute in } g 	imes 1000}{	ext{Equivalent weight} 	imes 	ext{Volume of solution in } mL}$.Given $10 \%$ (w/v) acetic acid means $10 \ g$ of acetic acid in $100 \ mL$ of solution.The molar mass of acetic acid $(CH_3COOH)$ is $60 \ g/mol$.Since the valency factor (n-factor) for acetic acid is $1$,the equivalent weight is equal to the molar mass,which is $60 \ g/eq$.Substituting the values: $N = \frac{10 	imes 1000}{60 	imes 100} = \frac{100}{60} = 1.666... \ N$.Rounding to one decimal place,we get $1.7 \ N$.

The normality of $10 \%$ (weight/volume) acetic acid is ......... $N$.

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