During a water treatment program, 127 grams of calcium nitrate, ca(no3)2, is dissolved in water. the final volume of the solution is 2,300 milliliters. what is the molarity of the solution?

Answer is:  the molarity of the solution is 0.336 M.

m(Ca(NO₃)₂) = 127 g; mass of calcium nitrate.

n(Ca(NO₃)₂) = m(Ca(NO₃)₂) ÷ M(Ca(NO₃)₂).

n(Ca(NO₃)₂) = 127 g ÷ 164.1 g/mol.

n(Ca(NO₃)₂) = 0.77 mol; amount of calcium nitrate.

V(solution) = 2300 mL ÷ 1000 mL/L.

V(solution) = 2.3 L.

c(Ca(NO₃)₂) = n(Ca(NO₃)₂) ÷ V(solution).

c(Ca(NO₃)₂) = 0.77 mol ÷ 2.3 L.

c(Ca(NO₃)₂) = 0.336 mol/L; molarity of the solution.

Data given:

                                                      mass of Ca(NO3)2 = 127 g

                                    molar mass of calcium nitrate = 164.08 g/mol

                                            moles of calcium nitrate = 127/164.08 = 0.77

                                                   Volume of solution = 2300ml =2.3 L

Solution

b.final