Question

$6.3$gram of oxalic acid, $0.4$ mole of oxalic acid and $0.2$ gram equivalents of oxalic acid are dissolved in a $2$ litre solution. $50$ml of this solution neutralises $20$ml of $KMn{O}_4$ solution in acidic medium. Molarity of $KMn{O}_4$ solution is:

• A. $0.285$M
• B. $0.550$M
• C. $0.137$M
• D. $0.350$M

Answer 1

The correct option is A $0.285$M

$1.$ Moles of $6.3$ $gm$ oxallic acid $=\frac{6.3}{90}=0.07$ $moles$

$2.$ Moles of $0.2$ $gram$ equivalents

$⟹0.2=Moles\times n_{factor}$

$⟹Moles=0.1$

$3.$ $+0.4$ $moles$

Total Moles $=0.1+0.4+0.07=0.57$ $moles$

Molarity $=\frac{0.57}{2}=0.285$ $M$

$Mn{O}_4^{-}+H_2{C}_2{O}_4⟶M{n}^{+2}+C{O}_2$

Equivalent of $Mn{O}_4^{-}=$ Equivalent of $C{O}_2$

$⟹20\times 5\times M_{KMn{O}_4}=50\times 0.285\times 2$

$M_{KMn{O}_4}=0.285$ Molar