A solution of Na_2S_2O_3 is standardized iodo | vedclass

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(A) The reaction involves the reduction of $BrO_3^-$ to $Br^-$. The change in oxidation state of $Br$ is from $+5$ to $-1$,so the $n$-factor for $KBrO

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$\frac{2}{15} \ N$

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(A) The reaction involves the reduction of $BrO_3^-$ to $Br^-$. The change in oxidation state of $Br$ is from $+5$ to $-1$,so the $n$-factor for $KBrO_3$ is $6$.Equivalents of $KBrO_3 = \frac{	ext{mass}}{	ext{Equivalent mass}} = \frac{0.167}{167/6} = \frac{0.167 	imes 6}{167} = 0.001 	imes 6 = 0.006 \ eq$.Since equivalents of $KBrO_3$ = equivalents of $Na_2S_2O_3$,we have:$N_1 V_1 = N_2 V_2$$0.006 = N 	imes 45 	imes 10^{-3} \ L$$N = \frac{0.006}{0.045} = \frac{6}{45} = \frac{2}{15} \ N$.

$A$ solution of $Na_2S_2O_3$ is standardized iodometrically against $0.167 \ g$ of $KBrO_3$ where $BrO_3^-$ changes to $Br^-$. This process requires $45 \ mL$ of the $Na_2S_2O_3$ solution. What is the strength of the $Na_2S_2O_3$ in $N$?
$[Mw \text{ of } KBrO_3 = 167]$

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$A$ solution of $Na_2S_2O_3$ is standardized iodometrically against $0.167 \ g$ of $KBrO_3$ where $BrO_3^-$ changes to $Br^-$. This process requires $45 \ mL$ of the $Na_2S_2O_3$ solution. What is the strength of the $Na_2S_2O_3$ in $N$?$[Mw \text{ of } KBrO_3 = 167]$