0.222 g of iron ore was brought into solution, Fe^(3+) is reduced to Fe^(2+) with SnCl_(2). The reduced solution required 20 mL of 0.1 N KMnO_(4) solution. The percentage of iron present in the ore is (equivalent weight of iron is 55.5 ) :

0.222 g of iron ore was brought into solution, Fe^(3+) is reduced to F

1.	0.222 g of iron ore was brought into solution, Fe^(3+) is reduced to Fe^(2+) with SnCl_(2). The reduced solution required 20 mL of 0.1 N KMnO_(4) solution. The percentage of iron present in the ore is (equivalent weight of iron is 55.5 ) :
Answer» <html><body><p>`55.5 %`<br/>`45.0%`<br/>`50.0 %`<br/>`40.0 %`</p>Solution :Number of <a href="https://interviewquestions.tuteehub.com/tag/equivalent-446407" style="font-weight:bold;" target="_blank" title="Click to know more about EQUIVALENT">EQUIVALENT</a> of <a href="https://interviewquestions.tuteehub.com/tag/iron-1051344" style="font-weight:bold;" target="_blank" title="Click to know more about IRON">IRON</a><br/> =Number of equivalents of `KMnO_(4)` <br/> `(<a href="https://interviewquestions.tuteehub.com/tag/nv-582326" style="font-weight:bold;" target="_blank" title="Click to know more about NV">NV</a>)/(1000)=(0.1xx20)/(1000)=0.002` <br/> Mass of iron (pure)`=0.002xx55.5=0.111 <a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a>` <br/> % purity `=("Mass of pure iron")/("Mass of iron ore")xx100=(0.111)/(0.222)xx100=50%`</body></html>