Answer:- [tex]9.50*10^1^8atoms[/tex] of Pb-204.
Solution:- From given info, the percent natural abundance of Pb-204 is 1.4%. It means, 1.4 grams of Pb-204 are present in 100 grams sample. With the help of this, we could calculate the amount of Pb-204 present in 230 mg that is 0.23 grams sample as:
[tex]0.23gSample(\frac{1.4gPb^2^0^4}{100gSample})[/tex]
= 0.00322 g of Pb-204
Let's convert grams to moles and then atoms as:
[tex]0.00322gPb(\frac{1mol}{204g})(\frac{6.02*10^2^3atoms}{1mol})[/tex]
= [tex]9.50*10^1^8atoms[/tex] of Pb-204
So, there would be [tex]9.50*10^1^8atoms[/tex] of Pb-204 present in the sample.
