Question

A 210 mL sample of an aqueous solution contains 3.15 %MgCl2 by mass (molar mass of MgCl2 is 96.9 g/mol). Exactly one-half of the magnesium ions are Mg−28, a beta emitter with a half-life of 21 hours. What is the decay rate of Mg−28 in the solution after 4.00 days? (Assume a density of 1.02 g/mLfor the solution.)

Answer 1

Mass of 210 ml of aqueous Mg CL2 solution =( 210 x 1,02 ) gms.

................................................................... = 214.20 gms   

Mass of Mg Cl2 in the given Mg Cl2 solution = ( 214.20 x 3.15 ) / 100

..................................................................... = 6.74 gms.

Number of moles of Mg Cl2 = ( 6.74 / 96.9 )

........................................... = 0.0696 moles

Mg Cl2 ionizes according to equation,

.......................................................Mg Cl2 (aq ) -----------> Mg ²⁺   + 2Cl^-      

stoichiometry ; ............................ 1 mole.............................1 mole.............2 moles...........

...................................................0.0696 moles..................0.0696 moles

Now, beta emitter magnesium ions produced = 1/2 ( 0.0696 )

........................................................................ = 0.0348 moles

Given T_1/2  of Mg-28 = 21 hours

so rate constant , k for decay of Mg-28 = 0.0696 / T_{1 / 2}  

................................................................ = 0.693 / 21

................................................................    = 0.033 hr^-1  

Radioactive decay follows ' first order kinetics ' , hence apply the relation

....................................................................................k = ( 2.303 / t ) log ( N_o / N )

where t is the given time period in hours = ( 24 x 4 ) = 96 hours , N_o  = initial number of moles of Mg-28 ion ;

N = number of moles of Mg-28 ions at given time

Substituting the given values we get, 0.033 = ( 2.303 / 96 ) log( 0.0348 / N )

hence N = 0.0015 moles

decay rate of Mg - 28 = ( N_o  - N ) / 96

..................................... = 3.47 x 10^-4  moles per hour

..............................or,.... = 8.325 x 10^-3 moles per day

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