Answer
$10.5 g/mL$
Work Step by Step
Given:
Weight of Silver (Ag): 194.3 g
Volume of Water (before the Silver is inserted), V${initial}$: 242.0 mL
Volume of Water (after the Silver is inserted), V${final}$: 260.5 mL
Formula:
Volume changed of Silver, $\Delta$V = V${final}$ - V${initial}$
$\Delta$V= 260.5mL - 242.0mL
$\Delta$V=18.5 mL
Density, ρ = $\frac{Mass, g}{Volume, mL}$
ρ $=\frac{194.3g}{18.5mL}$ $\approx10.5027027$
ρ $=10.5 g/mL$
