Make sure to always use liters with this equation. However, the unit g/mol is what is actually used in a calculation.Įxample #9: 10.0 g of acetic acid (CH 3COOH) is dissolved in 500. The M is the symbol for molarity and is used in discussing molarity. In the actual calculation, use mol/L rather than M. This next example is the most common type you'll see:Įxample #8: How many grams of KMnO 4 are needed to make 500. Remember it!!!Įxample #7: When 2.50 grams of KMnO 4 (molar mass = 158.0 g/mol) is dissolved into 100. Since the moles of the first equation equals the moles of the second equation, I can write this: First, I'll rearrange the second equation (with M for molarity and V for volume): The two steps just mentioned can be combined into one equation. To solve it, you convert mass (in grams) to moles, then divide by the volume, like this: What is the molarity of _ grams of dissolved in _ mL (or L) of solution. The most typical molarity problem looks like this: "A one molar solution is prepared by adding one mole of solute to sufficient water to make one liter of solution." It should be "one liter of solution" not "one liter of water." "A one molar solution is prepared by adding one mole of solute to one liter of water." The molarity definition is based on the volume of the solution, NOT the volume of pure water used. Notice how the phrase "of solution" keeps showing up. g/mol) is dissolved in enough water to make 1.00 L of solution. Step Two: dividing 1.00 mol by 2.00 L gives 0.500 mol/L (or 0.500 M).Ĭomment: remember that sometimes, a book will write out the word "molar," as in 0.500-molar.Įxample #5: Calculate the molarity of 25.0 grams of KBr dissolved in 750.0 mL.Įxample #6: 80.0 grams of glucose (C 6H 12O 6, mol. Step One: dividing 58.44 grams by 58.44 grams/mol gives 1.00 mol. The term "molar mass" is a moe generic term.) To solve the problem: (There is the term "formula weight" and the term "molecular weight." There is a technical difference between them that isn't important right now. In the above problem, 58.44 grams/mol is the molar mass of NaCl. Step Two: divide moles by liters to get molality. Eventually, the two steps will be merged into one equation. There two steps to the solution of this problem. ![]() What would be the molarity of the solution? All samples from a well-mixed solution will show the same concentration when analyzed.Įxample #4: Suppose you had 58.44 grams of NaCl and you dissolved it in exactly 2.00 L of solution. The solute has been completely dispersed throughout the entire extent of the solution. After all, chemists use balances to weigh things and balances give grams, NOT moles.īy the way, here's another point: solutions are always considered to be fully-mixed. Now, let's change from using moles to grams. And that number is 6.022 x 10 23 units, called Avogadro's Number.Įxample #3: What is the molarity when 0.750 mol is dissolved in 2.50 L of solution? One mole of sucrose or sodium chloride or anything else contains the same number of chemical units. It doesn't matter if it is sucrose, sodium chloride or any other substance. Notice that no mention of a specific substance is mentioned at all. The M is the symbol for molarity, the mol/L is the unit used in calculations.Įxample #2: Suppose you had 2.00 moles of solute dissolved into 1.00 L of solution. A dash is usually used when you write the word 'molar.'Īnd never forget this: replace the M with mol/L when you do calculations. By the way, you sometimes see 1.00 M like this: 1.00-molar. 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). When you handwrite it a block capital M is just fine. Some textbooks make the M using italics and some put in a dash, like this: 1.00- M. So, writing 1.00 M for the answer is the correct way to do it. Neither cancels.Ī symbol for mol/L is often used. Notice that both the units of mol and L remain. What would be the molarity of this solution? We then made sure that when everything was well-mixed, there was exactly 1.00 liter of solution. We keep adding water, dissolving and stirring until all the solid was gone. This is probably easiest to explain with examples.Įxample #1: Suppose we had 1.00 mole of sucrose (its mass is about 342.3 grams) and proceeded to mix it into some water. ![]() The molarity of a solution is calculated by taking the moles of solute and dividing by the liters of solution. As should be clear from its name, molarity involves moles.
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