Question

What mass of gold contains twice as many atoms as 1.59g of silver?

Answer 1

m(Ag) = 1.59 g

n(Ag) = m/M = 1.59 g / 107.87 g/mol = 0.01474 mol

N(Ag) = n*NA = 0.01474mol*6.02·1023atoms/mol = 8.873·1021 atoms

N(Au) = 2*8.873·1021 = 17.746·1021 atoms

n(Au) = N/NA = 17.746·1021 atoms / 6.02·1023 atoms/mol = 0.02948 mol

m(Au) = n*M = 0.02948 mol*196.97 g/mol = 5.81 g