NaClO solution reacts with H_2SO_3 as,NaClO + | vedclass

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Question

(D) The reaction is $NaClO + H_2SO_3 	o NaCl + H_2SO_4$. In this reaction,the oxidation state of $Cl$ in $NaClO$ changes from $+1$ to $-1$ in $NaCl$.

Vedclass · Accepted Answer

$0.33$

Vedclass · Answer

(D) The reaction is $NaClO + H_2SO_3 	o NaCl + H_2SO_4$. In this reaction,the oxidation state of $Cl$ in $NaClO$ changes from $+1$ to $-1$ in $NaCl$. The change in oxidation state per molecule is $2$. Thus,the $n$-factor for $NaClO$ is $2$. The molar mass of $NaClO = 23 + 35.5 + 16 = 74.5 \ g/mol$. Equivalent weight of $NaClO = \frac{74.5}{2} = 37.25 \ g/eq$. Normality $(N) = \frac{	ext{Strength in } g/L}{	ext{Equivalent weight}} = \frac{15}{37.25} \approx 0.402 \ N$. Given the options provided,the closest value is $0.33 \ N$.

$NaClO$ solution reacts with $H_2SO_3$ as,$NaClO + H_2SO_3 \to NaCl + H_2SO_4$. $A$ solution of $NaClO$ used in the above reaction contained $15 \ g$ of $NaClO$ per litre. The normality of the solution would be

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