Isotopic mass to density
This is possible, just use the density equation for a crystal lattice 
$$ d = \frac{z\times M}{V_c \times N_\mathrm A}$$ 
Where $d$ is density; $z$ is the number of atoms per unit cell; $M$ ist he molar mass of the material (isotope in this case); $V_c$ is the volume of a unit cell; and $N_\mathrm A$f is avogadro's number.  Since silver is FCC, $z= 4$ ($\frac{1}{8}\times 8 + \frac{1}{2}\times 6 = 4$) and $V_c$ is equal to the lattice parameter squared ($a^3; a = \pu{0.409nm}$).

This method will give you a very close approximate value but is limited in accuracy to the experimentally observed density as it does not account for free space in the crystal due to vacancies in the lattice or voids at grain boundaries. For example, if you calculated the densities of iridium and osmium you find that iridium should be denser, but experimentally you will not observe this result because iridium has a sufficiently high concentration of vacancies that osmium is actually denser.