Question

What volume (in mL) of 0.168 M BaCl2 is needed to get 5.08 x 10^22 chloride ions (Cl ̄)?

Answer 1

\[5.08*10^{22} * ({1 mol}/{6.022*10^{23}}) = 0.084367 moles of Chlorine ions\]
\[0.084357 mol* (1 mol BaCl_2/2mol Cl) = 0.042179 mol BaCl_2\]
\[0.042179 mol BaCl_2 * (1mol/0.168L) = 0.25 L, or 250 mL\]