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.)

A solution of 1 litre has 0.6g of non-radioactive Fe^(3+) with mass no

1.	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» SOLUTION :`{:(,.^(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",[0.209/57-10^(-5)],[0.6/56-10^(-5)],10^(-5),10^(-5)):}` `.^(57)Fe^(2+)` left after reaction `= 3.667xx10^(-3) - 10^(-5)` `= 366.6xx10^(5)` mole or `N_(0) = 366.6xx10^(-5)xx6.023xx10^(23) = 2.208xx10^(21)` Also `t = (2.303)/(lambda) log (N_(0))/(N)` `1 = (2.303xx4.62)/(0.693) log (2.208xx10^(21))/(N)` `N = 1.9xx10^(21)` `:.` Rate of decay `= lambda xx N = (0.693)/(4.62) xx 1.9xx10^(21)` `= 2.85xx10^(20)` DPH