Question

What is the molarity of phosphoric acid in a solution labelled 20.0% phosphoric acid (H3PO4) by weight   with a density = 1.12 g/mL?​

Answer 1

Lets recall the concept of molarity for solving this numerical. Molarity is defined as the number of moles of solute per liter of solution.

Molarity=$\frac{number of moles}{volume of solution (liter) }$

             =$\frac{n}{V}$

(n) number of moles=$\frac{given mass of solute}{molecular mass of solute}$

given values

given mass of solute  =20 gm

density= 1.12 g/ml

molecular mass of solute ($H_{3} PO_{4}$)= 1×3+31×1+16×4

                                                        = 98

(n) number of moles =$\frac{20}{98}$

                                  =0.204

volume of the solution= $\frac{given mass}{density}$

                                    = $\frac{20}{1.12}$

                                    = 17.85 ml

molarity = $\frac{number of moles}{volume of solution}$

             = $\frac{0.204}{17.85}$ ×$\frac{1000}{1}$

             =11.85 mol/L.

final answer is 11.85 mol/L