Answer to Question #105175 in General Chemistry for maria zal

Question

how many acetate ions are present in 18.0 mL of 0.688 M Pb(C2H3O2)2 solution?

Answer 1

First we will calculate moles of lead acetate =$M\times V =0.688\times18\times10^{-3}=12.4\times10^{-3}$

As 1 mole of lead acetate contains two moles of acetate ions

so $12.4\times10^{-3} moles$ will contain $2\times12.4\times10^{-3}$ N_A ions

Total number of acetate ions = $2\times12.4\times10^{-3}\times6.022\times10^{23}$ =$1.49\times10^{22}$