Question

For each of the following HCl solutions, the moles or grams of solute present and the total solution volume are given. Calculate the molarity of each solution. a.-1.40 moles HCl, 1.45 L solution b. 0.850 mole HCl, 0.867 L solution c. 30.0 g HCI, 1.45 L solution d. 30.0 g HCI, 875 mL solution 8-53 For each of the following KOH solutions, the moles or 8-54 re

Answer 1

8-53.

a.

Moles of HCl = 1.40 mol

Volume of HCl solution = 1.45 L

Molarity = moles of solute/liter of solution

              = 1.40 mol/1.45 L

              = 0.966 M

b.

Moles of HCl = 0.850 mol

Volume of HCl solution= 0.867 L

Molarity = moles of solute/liter of solution

              = 0.850 mol/0.867 L

              = 0.980 M

c.

Mass of HCl = 30.0 g

Molar mass of HCl = 36.5 g/mol

Moles of HCl = mass of HCl/molar mass of HCl

                      = 30.0 g/36.5 g/mol

                      = 0.822 mol

Moles of HCl = 0.822 mol

Volume of HCl solution= 1.45 L

Molarity = moles of solute/liter of solution

              = 0.822 mol/1.45 L

              = 0.567 M

d.

Mass of HCl = 30.0 g

Molar mass of HCl = 36.5 g/mol

Moles of HCl = mass of HCl/molar mass of HCl

                      = 30.0 g/36.5 g/mol

                      = 0.822 mol

Moles of HCl = 0.822 mol

Volume of HCl solution= 875 mL = 0.875 L     ( $\because$ 1 mL = 0.001 L)

Molarity = moles of solute/liter of solution

              = 0.822 mol/0.875 L

              = 0.939 M