S4E3 - Molarity Calculations. How to Find Molarity of the Solution

The extent to which a solute dissolves in a solution is expressed by the solution's concentration.

The Molarity Formula

Concentration is most often expressed as molarity (M).

Molarity (M)  =  number of moles of solute (mol)  /  liters of solution (L)

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The best way to understand how to use the molarity formula is to practice, practice, practice.  So let's dive in to some molarity examples now.

Ready?  Here we go...

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How to Find Molarity

ex:  Calculate the molarity of a solution made by dissolving 11.5 g of solid NaOH in enough water to make 1.50 L of solution.
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answer:

molarity  =  moles NaOH  /  L solution

➞  we already have the "L solution"  ➞  1.50 L

➞  but we need to calculate the "moles NaOH" before we can find the molarity...

Okay, so now we have both our numerator, 0.288 mol NaOH, and we always had our denominator (given), 1.50 L.  Let's divide the two and solve for molarity:

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ex:  Give the concentration of each type of ion in a solution of 0.50 M Co(NO₃)₂
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answer:

Think mole ratios.

For each 1 mole of cobalt (II) nitrate dissolved, the solution contains 1 mole Co²⁺ and 2 moles NO₃^-

So a solution that contains 0.50 M Co(NO₃)₂ contains:

0.50 M Co²⁺  and
1.0 M NO₃^- 

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ex:  Calculate the number of moles of chloride ions in 1.75 L of 1.0 M ZnCl₂
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answer:

NOTE - anytime you see [ X "L"  of  Y "M" ], you often have to multiply because  L^.M  =  L^.(mol/L)  =  mol

(1.75 L / 1) (1.0 mol ZnCl₂ / 1L)  =  1.75 mol ZnCl₂

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ex:  Blood serum is about 0.14 M NaCl.  What volume of blood contains 1.0 mg NaCl?
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answer:

If we convert 1.0mg NaCl to "moles NaCl," we can use the "0.14 moles / L" to get ourselves into units of liters (volume):

➞  (1.0mg NaCl / 1) (1g NaCl / 1000mg NaCl) (1mol NaCl / 58.45g NaCl)  =  1.7 x 10^-5 moles NaCl

➞  (1.7 x 10^-5 mol NaCl / 1) (1L solution / 0.14mol NaCl)  =  1.2 x 10^-4 L blood (solution)

Here's a better look at it...

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ex:  A chemist needs 1.00 L of an aqueous 0.200 M K₂Cr₂O₇ solution.  How much solid K₂Cr₂O₇ does he need to weigh out?
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answer:

We have [ X "L"  of  Y "M" ] ➞ L^.M  =  L^.(mol/L)  =  mol

➞  (1.00 L / 1) (0.200 mol K₂Cr₂O₇ / 1L)  =  0.200 mol K₂Cr₂O₇ 

➞  (0.200 mol K₂Cr₂O₇ / 1) (294 g K₂Cr₂O₇ / 1 mol K₂Cr₂O₇)  =  58.8g K₂Cr₂O₇

Here's an easy-to-follow version of the above two molarity calculations:

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Dilutions

People use the terms dilute and concentrated everyday to describe the contents of a liquid qualitatively.

dilute = "watered-down"
concentrated = contains "a lot" of ingredients that can be tasted

The Dilution Equation

From a quantitative point of view, chemists can prepare "dilutions" of solutions having an actual concentration.

Usually, highly concentrated "stock solutions" are purchased by a lab, and the chemists add water to achieve the desired molarity (concentration).

The Equation:   M₁V₁  =  M₂V₂

  --  M = molarity (mol / L)
  --  V = volume (matching units, often mL or L)

As stated in the above image, the moles of solute are not changed from situation 1 (M₁V₁) to situation 2 (M₂V₂).

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ex:  What volume of 16 M sulfuric acid must be used to prepare 1.5 L of a 0.10 M H₂SO₄ solution?
_________
answer:

Well, we have:

M₁ = 16 M     M₂ = 0.10 M
V₁ = ??             V₂ = 1.5 L

➞  M₁V₁ = M₂V₂

➞  (16 M) (x L)  =  (0.10 M) (1.5 L)

➞  (16x) mol  =  0.15 mol

➞  x = 0.0094 L or 9.4 mL

Here's a better look at all that...

To prepare 1.5 L of a 0.10 M H₂SO₄ solution, the chemist would add 9.4 mL of 16 M H₂SO₄ and then add enough water until the total final volume is 1.5 L.

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Next up in SECTION 4 - Types of Chemical Reactions and Solution Stoichiometry, we'll talk about the different Types of Chemical Reactions.