Molarity Dafination and Explanation
Molarity Problems
As should be clear from its name, molarity involves moles. Boy, does it!
The molarity of a solution is calculated by taking the moles of solute and dividing by the liters of solution.
Example #1: Suppose we had 1.00 mole of sucrose (it's about 342.3 grams) and proceeded to mix it into some water. It would dissolve and make sugar water. We keep adding water, dissolving and stirring until all the solid was gone. We then made sure that when everything was well-mixed, there was exactly 1.00 liter of solution.
What would be the molarity of this solution?
A symbol for mol/L is often used. It is a capital M. So, writing 1.00 M for the answer is the correct way to do it.
Some textbooks make the M using italics and some put in a dash, like this: 1.00-M. When you handwrite it; a block capital M is just fine.
When you say it out loud, say this: "one point oh oh molar." You don't have to say the dash (if it's there). By the way, you sometimes see 1.00 M like this: 1.00-molar. A dash is usually used when you write the word 'molar.'
And never forget this: replace the M with mol/L when you do calculations. The M is the symbol for molarity, the mol/L is the unit used in calculations.
Example #2: Suppose you had 2.00 moles of solute dissolved into 1.00 L of solution. What's the molarity?
Notice that no mention of a specific substance is mentioned at all. The molarity would be the same. It doesn't matter if it is sucrose, sodium chloride or any other substance. One mole of sucrose or sodium chloride or anything else contains the same number of chemical units. And that number is 6.022 x 1023 units, called Avogadro's Number.
Example #3: What is the molarity when 0.75 mol is dissolved in 2.50 L of solution?
Now, let's change from using moles to grams. This is much more common. After all, chemists use balances to weigh things and balances give grams, NOT moles.
Example #4: Suppose you had 58.44 grams of NaCl and you dissolved it in exactly 2.00 L of solution. What would be the molarity of the solution?
The solution to this problem involves two steps which will eventually be merged into one equation.
Example #5: Calculate the molarity of 25.0 grams of KBr dissolved in 750.0 mL.
Example #6: 80.0 grams of glucose (C6H12O6, mol. wt = 180. g/mol) is dissolved in enough water to make 1.00 L of solution. What is its molarity?
Notice how the phrase "of solution" keeps showing up. The molarity definition is based on the volume of the solution, NOT the volume of pure water used. For example, to say this:
Be careful on this, especially when you get to molality.
The most typical molarity problem looks like this:
What is the molarity of "whatever" grams of "whatever" substance dissolved in "whatever" mL of solution.
To solve it, you convert grams to moles, then divide by the volume, like this:
Example #7: When 2.00 grams of KMnO4 (molec. wt = 158.0 g/mol) is dissolved into 100.0 mL of solution, what molarity results?
This next example is the most common type you'll see:
Example #8: How many grams of KMnO4 are needed to make 500.0 mL of a 0.200 M solution?
Example #9: 10.0 g of acetic acid (CH3COOH) is dissolved in 500.0 mL of solution. What molarity results?
Example #10: How many mL of solution will result when 15.0 g of H2SO4 is dissolved to make a 0.200 M solution?
Molarity Problems
As should be clear from its name, molarity involves moles. Boy, does it!
The molarity of a solution is calculated by taking the moles of solute and dividing by the liters of solution.
This is probably easiest to explain with examples.
Example #1: Suppose we had 1.00 mole of sucrose (it's about 342.3 grams) and proceeded to mix it into some water. It would dissolve and make sugar water. We keep adding water, dissolving and stirring until all the solid was gone. We then made sure that when everything was well-mixed, there was exactly 1.00 liter of solution.
What would be the molarity of this solution?
The answer is 1.00 mol/L. Notice that both the units of mol and L remain. Neither cancels.
A symbol for mol/L is often used. It is a capital M. So, writing 1.00 M for the answer is the correct way to do it.
Some textbooks make the M using italics and some put in a dash, like this: 1.00-M. When you handwrite it; a block capital M is just fine.
When you say it out loud, say this: "one point oh oh molar." You don't have to say the dash (if it's there). By the way, you sometimes see 1.00 M like this: 1.00-molar. A dash is usually used when you write the word 'molar.'
And never forget this: replace the M with mol/L when you do calculations. The M is the symbol for molarity, the mol/L is the unit used in calculations.
Example #2: Suppose you had 2.00 moles of solute dissolved into 1.00 L of solution. What's the molarity?
The answer is 2.00 M.
Notice that no mention of a specific substance is mentioned at all. The molarity would be the same. It doesn't matter if it is sucrose, sodium chloride or any other substance. One mole of sucrose or sodium chloride or anything else contains the same number of chemical units. And that number is 6.022 x 1023 units, called Avogadro's Number.
Example #3: What is the molarity when 0.75 mol is dissolved in 2.50 L of solution?
The answer is 0.300 M.
Now, let's change from using moles to grams. This is much more common. After all, chemists use balances to weigh things and balances give grams, NOT moles.
Example #4: Suppose you had 58.44 grams of NaCl and you dissolved it in exactly 2.00 L of solution. What would be the molarity of the solution?
The solution to this problem involves two steps which will eventually be merged into one equation.
Step One: convert grams to moles.Step Two: divide moles by liters to get molality.In the above problem, 58.44 grams/mol is the molar mass of NaCl. (There is the term "formula weight" and the term "molecular weight." There is a technical difference between them that isn't important right now. The term "molar mass" is a moe generic term.) To solve the problem:
Step One: dividing 58.44 grams by 58.44 grams/mol gives 1.00 mol.Step Two: dividing 1.00 mol by 2.00 L gives 0.500 mol/L (or 0.500 M).Comment: remember that sometimes, a book will write out the word "molar," as in 0.500-molar.
Example #5: Calculate the molarity of 25.0 grams of KBr dissolved in 750.0 mL.
Example #6: 80.0 grams of glucose (C6H12O6, mol. wt = 180. g/mol) is dissolved in enough water to make 1.00 L of solution. What is its molarity?
Notice how the phrase "of solution" keeps showing up. The molarity definition is based on the volume of the solution, NOT the volume of pure water used. For example, to say this:
"A one molar solution is prepared by adding one mole of solute to one liter of water."is totally incorrect. It is "one liter of solution" not "one liter of water."
Be careful on this, especially when you get to molality.
The most typical molarity problem looks like this:
What is the molarity of "whatever" grams of "whatever" substance dissolved in "whatever" mL of solution.
To solve it, you convert grams to moles, then divide by the volume, like this:
The two steps just mentioned can be combined into one equation. Notice that moles is part of both equations, so one equation can be substituted into the other. Let's substitute the first into the second and rearrange just a bit to get this:
The M stands for molarity, the V for volume. In the second question, GMW has been substituted for molar mass. GMW stands for gram-molecular weight, which is a very common synonym for molar mass.
Example #7: When 2.00 grams of KMnO4 (molec. wt = 158.0 g/mol) is dissolved into 100.0 mL of solution, what molarity results?
This next example is the most common type you'll see:
Example #8: How many grams of KMnO4 are needed to make 500.0 mL of a 0.200 M solution?
Example #9: 10.0 g of acetic acid (CH3COOH) is dissolved in 500.0 mL of solution. What molarity results?
Example #10: How many mL of solution will result when 15.0 g of H2SO4 is dissolved to make a 0.200 M solution?
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