To determine the solubility of zinc cyanide you need the value of its solubility product constant, #K_(sp)#, which is listed at being #8.0 * 10^(-12)#.
So, zinc cyanide will not dissociate completely in aqueous solution; however, you can use the ICE table for the equilibrium reaction that takes place
#" "Zn(CN)2(s) rightleftharpoons Zn_((aq))^(2+) + color(red)(2)CN_((aq))^(-)#
I............#-#........................0...............0.
C..........#-#.......................(+x)..........(+#color(red)(2)#x)
E...........#-#........................x...............2x
Notice that, at equilibrium, you get 1 mole of #Zn^(2+)# cations and #color(red)(2)# moles of #(CN)^(-)# anions. The solubility product constant, #K_(sp)# for this reaction will be
#K_(sp) = [Zn^(2+)] * [CN^(-)]^(color(red)(2))#
#K_(sp) = x * (2x)^(2) = 4x^(3) = 8.0 * 10^(-12)#
Solve for #x# and you'll get
#x = color(green)(1.26 * 10^(-4)"mol/L")# #-># this is zinc cyanide's molar solubility.
Now, to get its solubility in grams per liter, all you have to do is multiply its molar solubility by its molar mass
#1.26 * 10^(-4)cancel("mol")/"L" * "117.44 g"/(1cancel("mole")) = color(green)(1.48 * 10^(-2)"g/L")#
