Question

28 ml of 0.1 m oxalic acid solution requires 10 ml of KMnO₄ for titration.10 ml of this sample of KMnO₄ when added to an excess of NH₂OH liberates N₂ at STP. Calculate the volume of N₂ liberated at STP?

Answer 1

Step 1: Given data

$28mlof0.1m$the oxalic acid solution requires $10ml$of ${\mathrm{KMnO}}_4$ for titration.

$10ml$ of this sample of ${\mathrm{KMnO}}_4$ when added to excess of ${\mathrm{NH}}_2\mathrm{OH}$ liberates ${\mathrm{N}}_2$ at STP.

Step 2 : Formula used

In acidic medium${\mathrm{MnO}}^{-4}\to {\mathrm{Mn}}^{+2}$

So $\left(\mathrm{NV}\right){}_{{\mathrm{KMnO}}_4}={\left(\mathrm{NV}\right)}_{{\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4}$
Step 3: Calculating Normality

Putting the values given above ,

$$\begin{gathered} N\times 10=28\times 0.1\times 2N \\ =0.56 \end{gathered}$$

Step 4: Calculating the Molarity of ${\mathrm{KMnO}}_4$ in acidic medium

M =$0.56/5\mathrm{with}{\mathrm{NH}}_2\mathrm{OH}$(medium is weakly basic).

$$\begin{gathered} {\left(\mathrm{NV}\right)}_{{\mathrm{KMnO}}_4}=\mathrm{W}/\mathrm{E}\mathrm{X}1000(0.56/5\mathrm{X}3)\mathrm{X}10 \\ =\frac{\mathrm{W}}{{\displaystyle \frac{\mathrm{M}}{2}}\mathrm{X}1000} \end{gathered}$$

Step 5: Calculating moles of${\mathrm{N}}_2$

=$1.68\times {10}^{-3}$

Step 6: Calculating Volume at STP

Volume =$1.68\times {10}^{-3}\times 22400$

$$\begin{gathered} =37.63 \\ =38ml. \end{gathered}$$

The volume of ${\mathrm{N}}_2$at STP= $38ml.$