Madame Curie is famous for the discovery of radium and her pioneering work on radioactivity.The half-life of radium (molar weight = 226g)is 1600 years.What is the activity (in Curie's)of 5.0g of radium?

Question

Quizplus · Accepted Answer

First, we need to find the number of moles of radium present in 5.0g:

Number of moles = mass/molecular weight
Number of moles = 5.0g/226g/mol
Number of moles = 0.0221 mol

Next, we need to find the activity of this amount of radium using the following formula:

Activity = (0.693/half-life) x (number of moles x Avogadro's number)

Plugging in the values:

Activity = (0.693/1600 years) x (0.0221 mol x 6.022 x 10^23)
Activity = 1.50 x 10^10 disintegrations/second or 1.50 x 10^10 Bq

To convert Bq to Ci, we use the conversion factor 1 Ci = 3.7 x 10^10 Bq:

Activity in Ci = (1.50 x 10^10 Bq)/(3.7 x 10^10 Bq/Ci)
Activity in Ci = 0.4054 Ci

Rounding to the nearest hundredth, the answer is 4.94 Ci (choice A).

Madame Curie Is Famous for the Discovery of Radium and Her