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A solution of 1 litre has 0.6g of non-radioactive Fe^(3+) with mass no. 56. To this solution 0.209g of radioactoveFe^(2+) is added with mass no. 57 and the following reaction occurred. .^(57)Fe^(2+) + .^(56)Fe^(3+) rarr .^(57)Fe^(3+)+ .^(56)Fe^(2+) At the end of one hour it was found that 10^(-5) moles of non-radioactive .^(56)Fe^(2+) mol L^(-1) hr^(-1). Negalecting any charge in volume, calculate the activity of the sample at the end of 1 hr (t_(1//2) for .^(57)Fe^(2+) = 4.62 hr.)

Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`{:(,.^(57)Fe^(2+)+,.^(56)Fe^(3+)to,.^(57)Fe^(3+)+,.^(56)Fe^(2+)),("Before reaction",0.209/57,0.6/56,0,0),("After reaction",[0.209/57-<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>^(-<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>)],[0.6/56-10^(-5)],10^(-5),10^(-5)):}` <br/>`.^(57)Fe^(2+)` left after reaction `= 3.667xx10^(-3) - 10^(-5)` <br/> `= 366.6xx10^(5)` mole <br/> or `N_(0) = 366.6xx10^(-5)xx6.023xx10^(23) = 2.208xx10^(<a href="https://interviewquestions.tuteehub.com/tag/21-293276" style="font-weight:bold;" target="_blank" title="Click to know more about 21">21</a>)`<br/> Also `t = (2.303)/(lambda) log (N_(0))/(N)` <br/> `1 = (2.303xx4.62)/(0.693) log (2.208xx10^(21))/(N)` <br/> `N = 1.9xx10^(21)` <br/> `:.` Rate of decay `= lambda xx N = (0.693)/(4.62) xx 1.9xx10^(21)` <br/> `= 2.85xx10^(20)` <a href="https://interviewquestions.tuteehub.com/tag/dph-2044953" style="font-weight:bold;" target="_blank" title="Click to know more about DPH">DPH</a></body></html>


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