To measure the quantity of MnCl_2 dissolved i | vedclass

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(D) The reaction involves the conversion of $Mn^{2+}$ to $MnO_4^-$ and then the titration of $MnO_4^-$ with oxalic acid $(H_2C_2O_4)$.Step $1$: Oxidat

Vedclass · Accepted Answer

$126$

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(D) The reaction involves the conversion of $Mn^{2+}$ to $MnO_4^-$ and then the titration of $MnO_4^-$ with oxalic acid $(H_2C_2O_4)$.Step $1$: Oxidation of $Mn^{2+}$ to $MnO_4^-$. The change in oxidation state of $Mn$ is from $+2$ to $+7$,so the n-factor is $5$.Step $2$: Reduction of $MnO_4^-$ by $H_2C_2O_4$. The change in oxidation state of $Mn$ is from $+7$ to $+2$ (n-factor $= 5$),and for $H_2C_2O_4$ to $CO_2$,the change in oxidation state of $C$ is from $+3$ to $+4$ (n-factor $= 2$ per molecule).By the law of equivalence,the number of equivalents of $MnCl_2$ equals the number of equivalents of $H_2C_2O_4$.$n_{MnCl_2} 	imes 5 = n_{H_2C_2O_4} 	imes 2$$\frac{w}{M_{MnCl_2}} 	imes 5 = \frac{225 \ mg}{90 \ g \ mol^{-1}} 	imes 2$Given $M_{MnCl_2} = 55 + 2 	imes 35.5 = 126 \ g \ mol^{-1}$.$\frac{w}{126} 	imes 5 = \frac{225}{90} 	imes 2$$\frac{w}{126} 	imes 5 = 2.5 	imes 2 = 5$$w = 126 \ mg$.

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