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:

$Molarity (M) = Moles of Solute (n) / Volume of Solution (L)$

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 the expression: Mass = n × M_r
    - The molar mass (M_r) of CuSO₄ is calculated as:
      - M_r = (M_r)_cu + (M_r)_S + 4(M_r)_O
      - M_r = 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

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:

n_solute (stock solution) = n_solute (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:

M_s × V_s = M_d × V_d

- 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. (M_r 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

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

Solution

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:
  - M_Na2SO4 = n × M_r
  - M_Na2SO4 = 0.0075 × 142.04
  - M_Na2SO4 = 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:
- V_H2O = 250 - 5.0
- V_H2O = 245 mL

Molarity (College Board AP Chemistry)

Revision Note

Molarity

Molarity

Preparing a Solution

Dilution

Worked example

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Author: Oluwapelumi Kolawole