Question

Question 2 4 pts Find the molality of a solution of ZnSO4 whose density is 1.25 g/mL and whose molarity is 1.19 M. (Show ALL of your work to receive full credit. All units must be included in your work and the answer needs to be reported to the correct number of significant figures. Answer must follow work.) Upload Choose a File

Answer 1

given ₃ 1.19 m molarity of zusoq solution means [1.19 moles of Znso are present in 1 lit (1000 ml) Solution. aho given density of Solution we know that density = mass So, density of selution = mass of solution 엉 = 1.25 gimo volume volume solution 1:25 mass glme of solution, 1000 ml & mass of solution 1-25 X 1000 = 1250 gm molality we need mass of solvent : mass of mass of zusoa for mass solvent Solution s mass mol solvent =/1250.gm 16198161.47 gm solvent = 11250 – 19.2.14) gn mass zusla of Х mass * = no.4 moles of zusog molecular ? weight q 2nsoq mass solvent = 1057.86 bgm ملح = mass g zuso a =1-19 X 161-47 gm mol mo molecular weight = 161.47 of zusoq gulmel

molali 5 lality no. of moles of zusoa mass x 1000 of solvent (gm) 1.19 moles = x 1000 1057.86 molality 1•12 mol/kg

By using given molarity and density values of solution we can calculate mass of solvent.

Then by putting values of number of moles of ZnSO4 and calculated mass of solvent in molality formula we can calculate molality of given solution.

Answer: molality = 1.12 mol/kg.