AP®ChemistryCollege BoardRevision NotesUnit 3: Properties of Substances & MixturesSolutions & MixturesMolarity

Molarity (College Board AP® Chemistry): Revision Note

Oluwapelumi Kolawole

Written by: Oluwapelumi Kolawole

Reviewed by: Stewart Hird

Updated on

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 by its concentration, which can be expressed in several ways; molarity is the most common in the laboratory and the one used on the AP exam

Molarity

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

Molarity (M)=moles of solute (n)volume of solution (L)

M=nV

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

  • Molarity is not the same as moles: moles of solute = molarity × volume in liters (n = M × V). Multiply, never divide, M by V

  • 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 × 0.2500 = 0.250 mol

    • Determine and weigh the corresponding mass of CuSO4 required

      • Using the expression: mass = n × molar mass

      • The molar mass of CuSO4 is calculated as:

      • molar mass = Cu + S + 4 O = 63.6 + 32.0 + 4(16.0) = 159.6 g/mol

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

    • Dissolve the 39.9 g of CuSO4 in a small volume of distilled water in a beaker, then transfer the solution to a 250.0 mL volumetric flask

    • Rinse the beaker with distilled water and add the rinsings to the flask, so no solute is lost

    • Add distilled water until the bottom of the meniscus sits on the calibration mark, then stopper and invert to mix

Preparing a solution

Steps 1 to 3 of making 250.0 mL of 1.00 M copper(II) sulfate solution. Step 1: blue solid on a balance reading 39.9 g, labeled 'Weigh out 39.9 g of CuSO4'. Step 2: the solid stirred with a glass rod in a beaker of water, labeled 'Add to a small volume of water and pre-dissolve the solid'. Step 3: the solution poured through a funnel into a volumetric flask, labeled 'Transfer to a volumetric flask'.
Step 4: the beaker rinsed with distilled water from a wash bottle into the flask, labeled 'Rinse the beaker with distilled water and add the rinsings to the flask'. Step 5: the stoppered flask with a magnified view of the neck showing the bottom of the meniscus on the calibration mark, labeled 'Make up to the calibration mark with more water. Add stopper and mix the contents'

How to prepare 250.0 mL of a 1.00 M CuSO4 solution

Dilution

  • Alternatively, a more concentrated solution, called the stock solution, can be diluted with water to give a solution of the desired molarity

    • Adding solvent cannot change the number of moles of solute, so:

nsolute (stock solution) = nsolute (dilute solution)

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

Examiner Tips and Tricks

  • Vd is the total volume of the diluted solution, not the volume of water added

    • The water added is Vd − Vs

  • Check your answer:

    • The stock volume Vs must be smaller than Vd (you cannot dilute 143 mL of stock to 100 mL)

  • Use Ms × Vs = Md × Vd only for dilution

    • For titrations and reactions, find moles with n = M × V and use the mole ratio from the balanced equation

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 (molar mass of glucose = 180.2 g/mol)

Answer:

Analyze: The mass of glucose and the volume of solution are given; find 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 liters

  • n = mass / molar mass

  • n = 5.00 / 180.2

  • n = 0.0277 mol

  • Volume = 100 / 1000

  • Volume = 0.100 L

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

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

  • Molarity (M) = 0.0277 / 0.100

  • Molarity (M) = 0.277 M

Worked Example

a) How many grams of Na2SO4 are there in 15 mL of 0.50 M Na2SO4?

b) How many milliliters of 0.50 M Na2SO4 solution are needed to provide 0.038 mol of this salt?

Analyze:

  • Part a gives the volume and molarity of the solution; find the mass of Na2SO4

  • Part b gives the number of moles of solute and the molarity of the solution; find the volume of solution in milliliters

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 liters

  • Convert to milliliters using the appropriate conversion factor

Answer:

  • Part a:

    • Step 1: Convert volume in mL to L:

      • Volume of solution = 15 / 1000 = 0.015 L

    • Step 2: Rearrange the molarity expression to find the moles of Na2SO4 (n):

      • n = M × V

      • n = 0.50 × 0.015

      • n = 0.0075 mol

    • Step 3: Convert the moles of solute to mass:

      • mass of Na2SO4 = n × molar mass

      • mass of Na2SO4 = 0.0075 × 142.04

      • mass of Na2SO4 = 1.1 g

  • Part b:

    • Step 1: Rearrange the molarity expression to find 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 milliliters 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?

Analyze:

  • The volume and concentration of the dilute solution are given; find the volume of the stock solution needed

  • Then find the volume of water needed to make the dilute solution

Plan:

  • Use the dilution formula to find the volume of stock solution

  • Then subtract that volume from the volume of the dilute solution to find the volume of water needed

Answer:

  • Step 1: Use Ms × Vs = Md × Vd

    • Rearrange the equation to Vs = (Md × Vd) / Ms

      • Vs = ?

      • Ms = 5.0 M

      • Vd = 250 mL

      • Md = 0.10 M

    • So, Vs = (0.10 × 250) / 5.0

    • Vs = 5.0 mL

  • Step 2: The volume of water added is:

    • Vwater = 250 − 5.0

    • Vwater = 245 mL

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

I would just like to say a massive thank you for putting together such a brilliant, easy to use website.I really think using this site helped me secure my top gradesin science and maths. You really did save my exams! Thank you.

Beth
IGCSE Student

This website is soooo useful and I can’t ever thank you enough for organising questions by topic like this. Furthermore, the name of the website could not have been more appropriate as it literally did SAVE MY EXAMS!

Fathima
A Level Student

Incredible! SO worth my money, the revision notes have everything I need to know and are so easy to understand. I actually enjoy revising! It makes me feel a lot more confident for my GCSEs in a few months.

Kate
GCSE Student

Absolutely brilliant, both my girls used it for A levels and GCSE. It's saves on paper copies, also beneficial exam questions ranked from easy to hard. It's removed a lot of stress from the exams.

Sameera
Parent

Just to say that your resources are the best I have seen and I have been teaching chemistry at different levels for about 40 years

Mark
Chemistry Teacher

Excellent

Our rating represented in Trustpilot stars
Read more reviews

Build on this topic

Check your understandingTry an Exam Question on this topicTest your recallReinforce key facts and definitions with Flashcards
Previous:Homogeneous & Heterogeneous MixturesNext:Particulate Representations of Solutions
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

Reviewer: Stewart Hird

Expertise: Chemistry Content Creator

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.