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

Sep 18, 2017

Approx. $500 \cdot g$.............
$\text{Moles of boron} = \frac{50.6 \cdot g}{10.81 \cdot g \cdot m o {l}^{-} 1} = 4.68 \cdot m o l$
Now remember that the mole is simply a number, i.e. $6.022 \times {10}^{23} \cdot m o {l}^{-} 1$... 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, $107.9 \cdot g \cdot m o {l}^{-} 1$.....
And so, 107.9*g*mol^-1xx4.68*mol-=??