1.1 g of sample of copper ore is dissolved and Cu^(2+) (aq) is treated with KI the iodine thus liberated requried 12.12 cm^(3) of 0.1 M Na_(2)S_(2)O_(3) solution for tritration what is the percentage of copper in the ore ?

1.1 g of sample of copper ore is dissolved and Cu^(2+) (aq) is treated

1.	1.1 g of sample of copper ore is dissolved and Cu^(2+) (aq) is treated with KI the iodine thus liberated requried 12.12 cm^(3) of 0.1 M Na_(2)S_(2)O_(3) solution for tritration what is the percentage of copper in the ore ?
Answer» <html><body><p></p>Solution :The complete balanced equation for the redox reaction is <br/> `2cu^(2+)+4I^(-)+2S_(2)O_(3)_^(2)rarrCu_(2)I_(2)+S_(2)O_(6)^(2-)+2I^(-)` <br/> No of <a href="https://interviewquestions.tuteehub.com/tag/moles-1100459" style="font-weight:bold;" target="_blank" title="Click to know more about MOLES">MOLES</a> of `S_(2)O_(3)^(2-)`used =`(12.12)/(<a href="https://interviewquestions.tuteehub.com/tag/1000-265236" style="font-weight:bold;" target="_blank" title="Click to know more about 1000">1000</a>)xx0.1=1.212xx10^(-3)` moles <br/> From the <a href="https://interviewquestions.tuteehub.com/tag/balance-891682" style="font-weight:bold;" target="_blank" title="Click to know more about BALANCE">BALANCE</a> equation <br/> 2 moles of `S_(2)O_(3)^(2-)` reduce `Cu^(2+)` =2 moles <br/> `therefore 1.212 xx10^(-3)` moles of `S_(2)O_(3)^(2-)` will reduce `Cu^(2+)=1.212 xx10^(-3)` moles <br/> Thus % <a href="https://interviewquestions.tuteehub.com/tag/age-851457" style="font-weight:bold;" target="_blank" title="Click to know more about AGE">AGE</a> of Cu in the are =`(0.77)/(1.1)xx100=7%`</body></html>