go to the AUS-e-TUTE homepage
home test Join AUS-e-TUTE game contact
 

 

Concentration of Solutions (Molarity) Calculations

Key Concepts

  • The concentration of a solution is usually given in moles per litre (mol L-1 OR mol/L).
        This is also known as molarity1.

  • Concentration in mol/L or mol L-1, or Molarity, is given the symbol c (sometimes M).
        For a 0.01 mol L-1 HCl solution we can write :
            [HCl] = 0.01 mol L-1 (concentration implied by square brackets around formula)
            or
            c(HCl) = 0.01 mol L-1 (c stands for concentration, formula given in brackets)

  • c = n ÷ V
        where
          c = concentration of solution in mol L-1 (mol/L or M),
          n = moles of substance being dissolved (moles of solute),
          V = volume of solution in litres (L)

  • This formula can be re-arranged to find:
          moles of solute given molarity and volume of solution: n = c x V
          volume of solution given moles of solute and molarity: V = n ÷ c

Examples

1. Calculating Concentration (c = n ÷ V)

Calculate the concentration in mol L-1 (molarity) of a sodium chloride solution containing 0.125 moles sodium chloride in 0.5 litres of solution.

  1. Extract the data from the question
        n = 0.125 mol
        V = 0.5 L

  2. Write the equation:
        c = n ÷ V

  3. Substitute in the values and solve:
        [NaCl(aq)] = c(NaCl(aq)) = 0.125 ÷ 0.5 = 0.25 mol L-1 (or 0.25 mol/L or 0.25 M)

2. Calculating Moles of Solute (n = c x V)

Calculate the moles of copper sulfate in 250 mL of 0.02 mol L-1 copper sulfate solution.

  1. Extract the data from the question:
        c = 0.02 mol L-1
        V = 250 mL = 250 ÷ 1000 = 250 x 10-3 L = 0.250 L (since there are 1000 mL in 1 L)

  2. Write the equation:
        n = c x V

  3. Substitute in the values and solve:
       n = 0.02 x 250 x 10-3 = 0.005 mol

3. Calculating Volume of Solution (V = n ÷ M)

Calculate the volume of a 0.80 mol L-1 potassium bromide solution containing 1.6 moles of potassium bromide.

  1. Extract the data from the question:
        n = 1.6 mol
        c = 0.80 mol L-1

  2. Write the equation:
        V = n ÷ c

  3. Substitute in the values and solve:
        V = 1.6 ÷ 0.80 = 2.00 L


What would you like to do now?

1 There are many other ways to measure concentration of solutions in chemistry. You will find tutorials dealing with these under the heading Solutions on the AUS-e-TUTE homepage.

advertise on the AUS-e-TUTE website and newsletters
 
 

Search this Site

You can search this site using a key term or a concept to find tutorials, drills, tests, exams, learning activities (games), worksheet and quiz wizards.
 

Become an AUS-e-TUTE Member

 

AUS-e-TUTE's Blog

 

Subscribe to our Free Newsletter

Email email us to
subscribe to AUS-e-TUTE's free quarterly newsletter, AUS-e-NEWS.

AUS-e-NEWS quarterly newsletter

AUS-e-NEWS is emailed out in
December, March, June, and September.

 

Ask Chris, the Chemist, a Question

The quickest way to find the definition of a term is to ask Chris, the AUS-e-TUTE Chemist.

Chris can also send you to the relevant
AUS-e-TUTE tutorial topic page.

 
 
 

Share this Page

Bookmark and Share
 
 

© AUS-e-TUTE