Question

Calculate the molarity of a solution that contains 3.00×10⁻² mol NH4Cl in exactly 430 mL of solution.

How many moles of HNO3 are present in 36.0 mL of a 1.80 M solution of nitric acid?

How many milliliters of 1.70 M KOH solution are needed to provide 0.125 mol of KOH?

Answer 1

(1) Given volume , V = 430 mL = 0.430 L

    Number of moles , n = 3.00x10^-2 mol

So Molarity , M = number of moles / volume in L

                        = (3.00x10^-2 mol) / 0.430 L

                       = 0.07 M

(2) Given volume , V = 36.0 mL = 36.0/1000 = 0.036 L

         MOlarity , M = 1.80 M

    Number of moles , n = ?

So Molarity , M = number of moles / volume in L

So number of moles , n = molarity x volume in L

                                  = 1.80 M x 0.036L

                                  = 0.065 mol

(3) The volume in L required is , V = Number of moles / molarity

                                                 = 0.125 mol / 1.70 M

                                                 = 0.0735 L

                                                 = 0.0735 x 1000 mL

                                                 = 73.5 mL