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The Molarity Formula and Molarity Calculations. How to Find Concentration of the Solution

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

Concentration is most often expressed as molarity (M).

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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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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:

— — — — —

Think mole ratios.

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

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

0.50 M Co^2+ and
1.0 M NO₃-

— — — — —

NOTE - anytime you see [ X “L” of Y “M” ], you often have to multiply because L*M = L*(mol/L) = moles

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

— — — — —

— — — — —

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 of blood (solution).

Here’s a better look at it…

— — — — —

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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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

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 Dilution Equation:

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).

— — — — —

Well, we have:

M₁ = 16 M
V₁ = ??

M₂ = 0.10 M
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...

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