Question

A 3.4 g sample of $H_2{O}_2$ solution containing $x\%H_2{O}_2$ by mass requires x mL of a $KMn{O}_4$ solution for complete oixidation under acidic condition.The molarity of $KMn{O}_4$ solution is:

• A. 1
• B. 0.5
• C. 0.4
• D. 0.2

Answer 1

The correct option is A 1

Molarity $:\to$ The moles of solute present in one litre ofsolution is known as molarity.

$\because 100gm$ of $KMn{O}_4$ solutions contain $xgm$ of $H_2{O}_2$

$\therefore 1gm$ of $KMn{O}_4$ solutions contain $\frac{x}{100}gm$ of $H_2{O}_2$

$\therefore 3.4gm$ ______ $\frac{x}{100}\times 3.4$ of $H_2{O}_2\left(gm\right)$

moles of $H_2{O}_2=\frac{x}{100}\times \frac{3.4}{34}(34=$ Mol wt of $H_2{O}_2)$

$=x\times {10}^{-3}moles$

$\therefore$ molarity $=\frac{molesof{H}_2{O}_2}{Volumeof{KMnO}_{4}solutionin\left(lt\right)}=\frac{x\times {10}^{-3}mole/lt}{xml\times {10}^{-3}}=1mol/lt$