Question

$40gBa(Mn{O}_4{)}_2$ (mol. wt. $=375$ g/mol) sample containing some inert impurities in acidic medium is completely reacted with $125$ mL of $33.6V$ of $H_2{O}_2$. What is the percentage purity of the sample?

• A. $28.12\%$
• B. $70.31\%$
• C. $85\%$
• D. None of the above

Answer 1

The correct option is B $70.31\%$

Mass of $H_2{O}_2$ in $125$ ml solution $=\frac{17}{5600}\times 33.6\times 125$ g
So, number of equivalents of $H_2{O}_2=33.6\times \frac{125}{5600}$
As number of equivalents of both are same,
$\frac{x}{\frac{375}{10}}=33.6\times \frac{125}{5600}$

or, $x=28.125$
So, percentage purity of the sample $=\frac{28.125}{40}\times 100=70.31\%$