Question

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 1

• The morality of solution is 0.33 M

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

• Molarity = No. of moles × 1/vol. of solution in litre

• M = 0.77 × 1/2.3 = 0.33 M

Answer 2

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.