AP®ChemistryCollege BoardStudy GuidesUnit 3: Intermolecular Forces & PropertiesSolutions & MixturesMolarity

Molarity (College Board AP® Chemistry)

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Oluwapelumi Kolawole

Written by: Oluwapelumi Kolawole

Reviewed by: Stewart Hird

Molarity

  • A solution may be described qualitatively or quantitatively

    • Qualitatively, the terms dilute and concentrated are used to describe the amount of solute in a given quantity of the solvent

      • A solution with a relatively small concentration of solute is said to be dilute

      • A solution with a large concentration of solute is said to be concentrated

    • Quantitatively, a solution is described in terms of its concentration

Molarity

  • Molarity (M) expresses the concentration of a solution as the number of moles of solute in a litre of solution:

Molarity space left parenthesis straight M right parenthesis space equals space Moles space of space Solute left parenthesis straight n right parenthesis divided by Volume space of space Solution left parenthesis straight L right parenthesis

M = n / V

  • A 1.00 molar solution (1.00 M) contains 1.00 mol of solute in every 1.0 L of solution

  • A solution can be prepared to a specified molarity by weighing out the calculated mass of solute and dissolving it in enough solvent to form the desired volume of solution

  • For example, to prepare 250.0 mL of a 1.00 M solution of CuSO4:

    • Determine the number of moles of the solute (CuSO4) required

      • In this case, this is given as n = Molarity × Volume

      • n = 1.00 × 0.250  = 0.250 mol

    • Determine and weigh the corresponding mass of CuSO4 required

      • Using the expression: Mass = n × Mr

      • The molar mass (Mr) of CuSO4 is calculated as:

      • Mr = (Mr)cu + (Mr)S + 4(Mr)O

      • Mr = 63.6 + 32.0 + 4 (16) = 159.6 g/mol

      • Mass = 0.250 mol × 159.6 g/mol = 39.9 g

      • Transfer 39.9 g of CuSO4 into a 250 mL flask and add some quantity of water to dissolve the solute

      • Add more water until the solution reaches the calibrated mark of the flask

Preparing a Solution

1-2-7-preparing-a-standard-solution-1
1-2-7-preparing-a-standard-solution-2-1

How to prepare a 250mL of 1.00 M solution of CuSO4

Dilution

  • Alternatively, you can start with a more concentrated solution, called the stock solution, and dilute it with water to give a solution of the desired molarity

    • The calculations are straightforward if you keep a simple point in mind: Adding solvent cannot change the number of moles of solute

      • That is:

nsolute (stock solution) =  nsolute (dilute solution)

  • In both solutions, n can be found by multiplying the molarity, M, by the volume in litres, V

  • Hence, a dilution expression can be obtained:

Ms × Vs = Md × Vd

  • Where the subscripts s and d stand for stock and dilute solutions, respectively

Worked Example

Calculate the molarity of a solution made by dissolving 5.00 g of glucose (C6H12O6) in sufficient water to form exactly 100 mL of solution. (Mr glucose = 180.2 g/mol)

Answer:

Analyse: We are provided with the mass of glucose and the volume of solution and asked to determine the molarity of the solution

Plan: To determine the molarity of the solution the following steps are required:

Step 1: Convert the mass of glucose to moles and the volume of the solution to litres

  • n = Mass/Molar Mass        

  • n = 5.00/180.2                    

  • n = 0.0277 moles

  • Volume = 100/1000

  • Volume = 0.1L

Step 2: Substitute the values of the moles of glucose and the volume of the solution

  • Molarity (M) = number of moles/volume of solution

  • Molarity (M) = 0.0277/0.1

  • Molarity (M) = 0.277 M

Worked Example

  1. How many grams of Na2SO4 are there in 15 mL of 0.50 M Na2SO4?

  2. How many millilitres of 0.50 M Na2SO4 solution are needed to provide 0.038 mol of this salt?

Analyse:

  • In part a, we are given the volume of the solution and the concentration (Molarity)

    • We are then asked to determine the mass of the solute(Na2SO4) required

  • In part b, we are given the number of moles of solute and the molarity of the solution

    • We are then asked to determine the volume of solution in millilitres

Plan:

For part a:

  • Use the molarity expression to determine the number of moles of the solute (n)

  • Convert this number of moles to mass

For part b:

  • Use the molarity expression to determine the volume of solution in litres

  • Convert to millilitres using the appropriate conversion factor

Answer:

  • Part a:

    • Step 1: Convert volume in mL to L:

      • Volume of solution = 15/1000 = 0.015L

    • Step 2: Rearrange the concentration expression in terms of the number of Na2SO4 (n):

      • n = M × V

      • n = 0.50 × 0.015

      • n = 0.0075 moles

    • Step 3: Using the mole-molar mass expression, convert the moles of solute to mass:

      • MNa2SO4 = n × Mr

      • MNa2SO4 = 0.0075 × 142.04

      • MNa2SO4 = 1.1g

  • Part b:

    • Step 1: Rearrange the concentration expression in terms of the volume of solution:

      • V = n/M

      • V = 0.038/0.50

      • V = 0.076 L

    • Step 2: Convert volume in L to mL:

      • V = 0.076 × 1000

      • V = 76 mL

Worked Example

How many millilitres of 5.0 M Na2Cr2O7 solution must be diluted to prepare 250 mL of 0.10 M solution? How much water needs to be added to this concentrated solution?

Analyse:

  • We are given the volume and concentration of the diluted solution and asked to determine the volume of the concentrated solution required to prepare

  • We are then asked the volume of water required to make the dilute solution

Plan:

  • Using the dilution formula, we can calculate the volume of the concentrated solution required

  • Then subtract that volume from the volume of the dilute solution to obtain the volume of water required

Answer:

  • Step 1: Using M1V1 = M2V2

    • Rearrange the equation to V1 = (M2V2)/ M1

      • V1 = ?

      • M1 = 5.0 M

      • V2 = 250 mL

      • M2 = 0.10 M

    • So, V1 = (0.10 × 250)/5.0

    • V1 = 5.0 mL

  • Step 2: The volume of water added will be:

    • VH2O = 250 - 5.0

    • VH2O = 245 mL

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    Author
    Reviewer
    Oluwapelumi Kolawole

    Author: Oluwapelumi Kolawole

    Expertise: Chemistry Content Creator

    Oluwapelumi is a Pharmacist with over 15000+ hours of AP , IB, IGCSE, GCSE and A-Level chemistry tutoring experience. His love for chemistry education has seen him work with various Edtech platforms and schools across the world. He’s able to bring his communication skills as a healthcare professional in breaking down seemingly complex chemistry concepts into easily understood concepts for students.

    Stewart Hird

    Author: Stewart Hird

    Expertise: Chemistry Lead

    Stewart has been an enthusiastic GCSE, IGCSE, A Level and IB teacher for more than 30 years in the UK as well as overseas, and has also been an examiner for IB and A Level. As a long-standing Head of Science, Stewart brings a wealth of experience to creating Topic Questions and revision materials for Save My Exams. Stewart specialises in Chemistry, but has also taught Physics and Environmental Systems and Societies.