Some Basic Concepts of Chemistry - Result Question 33
Subjective Questions
34. To measure the quantity of $\mathrm{MnCl}_2$ dissolved in an aqueous solution, it was completely converted to $\mathrm{KMnO}_4$ using the reaction,
$\mathrm{MnCl}_2+\mathrm{K}_2 \mathrm{~S}_2 \mathrm{O}_8+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{KMnO}_4+\mathrm{H}_2 \mathrm{SO}_4+\mathrm{HCl}$ (equation not balanced).
Few drops of concentrated HCl were added to this solution and gently warmed. Further, oxalic acid $(225 \mathrm{mg})$ was added in portions till the colour of the permanganate ion disappeared. The quantity of $\mathrm{MnCl}_2$ (in mg ) present in the initial solution is $\qquad$
(Atomic weights in $\mathrm{g} \mathrm{mol}^{-1}: \mathrm{Mn}=55, \mathrm{Cl}=35.5$ )
(2018 Adv.)
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Answer:
Correct Answer: 34. (126 mg)
Solution:
- The balanced equations are (1) $2 \mathrm{MnCl}_2+5 \mathrm{~K}_2 \mathrm{~S}_2 \mathrm{O}_8+8 \mathrm{H}_2 \mathrm{O} \rightarrow$
$2 \mathrm{KMnO}_4+4 \mathrm{~K}_2 \mathrm{SO}_4+6 \mathrm{H}_2 \mathrm{SO}_4+4 \mathrm{HCl}$
(2)
$\begin{aligned} & 2 \mathrm{KMnO}_4+5 \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4+3 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow \\ & \mathrm{K}_2 \mathrm{SO}_4+2 \mathrm{MnSO}_4+8 \mathrm{H}_2 \mathrm{O}+10 \mathrm{CO}_2 \\ \end{aligned}$
Given, mass of oxalic acid added $=225 \mathrm{mg}$
So, millimoles of oxalic acid added $=\frac{225}{90}=2.5$ Now from equation 2
Millimoles of $\mathrm{KMnO}_4$ used to react with oxalic acid=1 and Millimoles of $\mathrm{MnCl}_2$ required initially=1
$\therefore$ Mass of $\mathrm{MnCl}_2$ required initially $=1 \times(55+71)=126 \mathrm{mg}$
Alternative Method
$m$ moles of $\mathrm{MnCl}_2=m$ moles of $\mathrm{KMnO}4=x$ (let) and $M{\text {eq }}$ of $\mathrm{KMnO}4=M{\text {eq }}$ of oxalic acid
So, $\quad x \times 5=\frac{225}{90} \times 2$
Hence, $x=1$
$\therefore \quad m$ moles of $\mathrm{MnCl}_2=1$
Hence mass of $\mathrm{MnCl}_2=(55+71) \times 1=126 \mathrm{mg}$.
BETA
