Question

A. Calculate the molarity of a solution that contains 0.165 molZNCl2 in exactly 160 mL of solution.

B. How many moles of protons are present in 34.0mL of a 4.70 M solution of nitric acid?

C. How many milliliter of 6.20 M NaOh solutions are needed to provide 0.315 mol of NaOh?

Answer 1

A. no of moles of ZnCl2   = 0.165moles

volume of solution = 160ml   = 0.16L

molarity of ZnCl2   = no of moles/volume of solution in L

                             = 0.165/0.16

                              = 1.03125M

B. molarity of HNO3 = 4.7M

   volume of solution = 34ml = 0.034L

no of moles of HNO3 = molarity * volume of solution in L:

                                    = 4.7*0.034   = 0.1598moles

C. no of moles of NaOH   = 0.315M

    molarity of NaOH   = 6.2M

   no of moles of NaOH   = molarity *volume in L

                             0.315     = 6.2* volume of solution in L

volume of solution in L   = 0.315/6.2   = 0.0508L

                    volume of solution = 50.8ml