How to determine the specific activity of an experimental solution

Suppose you need to make up a certain volume of an experimental solution, to contain a particular amount of radioactivity. For example, 50 mL of a mannitol solution at a concentration of 25mmolL to contain 5Bq/(L - using a manufacturer s stock solution of C-labelled mannitol (specific activity = 0.1 Ci mmol h. 

Calculate the total amount of radioactivity in the experimental solution, in this example 5x1000 (to convert juL to ml) x 50 (50 ml required) = 2.5 X 10 Bq (i.e. 250kBq). 

Establish the volume of stock radioisotope solution required for example, a manufacturer s stock solution of C-iabelled mannitol contains 50/(Ci of radioisotope in 1 ml of 90% v/v ethanol water. Using Table 35.3, this is equivalent to an activity of 50 x 37 = 1 850kBq. So, the volume of solution required is 250/1850 of the stock volume, i.e. 0.135 IrnLd35/(L). 

Check the amount of radioactive isotope to be added. In most cases, this represents a negligible amount of substance, e.g. in this instance, 250 kBq of stock solution at a specific activity of 14,8x 10 kBqmmor (converted from 0 4Ci mmol using Table 35.3) is equal to 250/14800000 = 16.89 nmol, which is equivalent to approximately 3 /rg mannitol. This can be ignored in calculating the mannitol concentration of the experimental solution. 

Make up the experimental solution by adding the appropriate amount of non-radioactive substance and the correct volume of stock solution. 

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