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# amu/moles/number of atoms Watch

1. Calculate the number of atoms present in 1kg of 226 88 Ra?

So I did (88 x 1.00728)+(138 x 1.00867)= 227.8371 Amu/molar mass
then 227.8371 x 6.02x10^23 (avogadro's constant) to get 1.37..x10^26 which was wrong.

How do you do these types of question? Is there a better way to work out the number of atoms?
Thanks
2. (Original post by Lay-Z)
Calculate the number of atoms present in 1kg of 226 88 Ra?

So I did (88 x 1.00728)+(138 x 1.00867)= 227.8371 Amu/molar mass
then 227.8371 x 6.02x10^23 (avogadro's constant) to get 1.37..x10^26 which was wrong.

How do you do these types of question? Is there a better way to work out the number of atoms?
Thanks
So the mass of an atom of 226 Ra is 227.8371u (assuming your calc is correct). That is 3.782*10^(-25) kg. (I have multiplied by u).

So the number of atoms in 1kg is 1 /(3.782*10^(-25)) = 2.64*10^(24)

Alternatively, the number of moles is 1000/(227.8371) (by moles = mass/ mr) (remember to work in grams) and then you can multiply this number by avogradro's constant
3. thanks

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