Silver is fcc with edge length 409 pm and density 10.5 g cm^-3. Using N_A = 6.022x10^23 mol^-1, calculate its atomic mass (nearest integer, in g mol^-1).

JEE Main levelNumerical answerDifficulty 2 of 3ss066

Silver is fcc with edge length $409 pm$ and density $10.5 g c m^{- 3}$ . Using $N_{A} = 6.022 \times 1 0^{23} mo l^{- 1}$ , calculate its atomic mass (nearest integer, in $g mo l^{- 1}$ ).

Numerical answer type, answer in g/mol, tolerance 1.

M = \frac{ρ N _{A} a ^{3}}{Z}

with

Z = 4

M = \frac{10.5 \times 6.022 \times 1 0 ^{23} \times ( 4.09 \times 1 0 ^{- 8} ) ^{3}}{4} = 108.15 \approx 108