Question

What mass of silver (Ag) contains the same number of atoms as 50.6 g of boron (B)?

Answer 1

Approx. `500*g`.............

Explanation:

`"Moles of boron"=(50.6*g)/(10.81*g*mol^-1)=4.68*mol`

Now remember that the mole is simply a number, i.e. `6.022xx10^23*mol^-1`... And to get the same number of silver atoms, all I have to do is multiply this molar quantity by the molar mass of silver, `107.9*g*mol^-1`.....

And so, `107.9*g*mol^-1xx4.68*mol-=??`