Molarity, Dilution

Molarity  Molarity is the number of moles of solute dissolved in one liter of solution. The units, therefore are moles per liter, specifically it's moles of solute per liter of solution.  Rather than writing out moles per liter, these units are abbreviated as M or M. We use a capital M with a line under it or a capital M written in italics. So when you see M or M it stands for molarity, and it means moles per liter (not just moles).  You must be very careful to distinguish between moles and molarity. "Moles" measures the amount or quantity of material you have; "molarity" measures the concentration of that material. So when you're given a problem or some information that says the concentration of the solution is 0.1 M that means that it has 0.1 mole for every liter of solution; it does not mean that it is 0.1 moles. Please be sure to make that distinction. molarity = moles of solute liter of solution

Molarity calculations  There are several types of calculations that you need to be able to do with molarity. First, you should be able to calculate the molarity if you are given the components of the solution. Second, you should be able to calculate the amount of solute in (or needed to make) a certain volume of solution. Third, you might need to calculate the volume of a particular solution sample. Fourth, you might need to calculate the concentration of a solution made by the dilution of another solution.

Calculating Molarity from Moles and Volume  Here we are given something to figure out. To get the molarity we need to divide the number of moles of NaCl by the volume of the solution. In this case that is 0.32 moles NaCl divided by 3.4 L, and that gives M NaCl. What is the molarity of a solution containing 0.32 moles of NaCl in 3.4 liters? molarity = 0.32 moles NaCl 3.4 L =0.094 M NaCl

What is the molarity of a solution made by dissolving 2.5 g of NaCl in enough water to make 125 ml of solution? molarity = moles of solute liter of solution Calculating Molarity from Mass and Volume 2.5 g NaCl x 1 mole NaCl 58.5 g NaCl = mole molarity = mole NaCl L =0.34 M NaCl This one is a bit more difficult. To get molarity we still need to divide moles of solute by volume of solution. But this time we're not given the moles of solute. We have to calculate it from the mass of NaCl. We multiply 2.5 g NaCl by the conversion factor of 1 mole NaCl over the formula weight of NaCl, 58.5 g. That tells us that we have mole of NaCl. I kept an extra digit here because we are not done with the calculations. When we are done I'll round off to two digits, the same as in the 2.5 g weight of NaCl. Now that we know the moles we can calculate the molarity. Moles of solute (0.0427) divided by the volume of the solution (0.125 L) gives us 0.34 M NaCl. n=m/M n (amount of substance) = mass/molar mass Unit is 1 mol

How would you prepare 400. ml of 1.20 M solution of sodium chloride? 1.20 M NaCl = 1.20 moles NaCl 1.00 L solution L solution x 1.20 moles NaCl 1.00 L solution =0.480 moles NaCl moles NaCl x 58.5 g NaCl 1 mole NaCl =28.1 g NaCl Dissolve 28.1 g NaCl in enough water to make 400 mL of solution. Calculating Mass of Solute from Molarity Algebraic approach: Set up general equation: x M = y moles z L Substitute known values: 1.20 M = y moles L Solve for moles: y moles = 1.20 M x L = moles n=m/M n (amount of substance) = mass/molar mass Unit is 1 mol

Calculating Moles of Solute from Molarity How many moles of salt are contained in 300. mL of a 0.40 M NaCl solution? L x 0.40 moles NaCl 1.00 L solution =0.12 moles NaCl This question is a little easier. We do it the same way as the first step of the previous problem and then we stop. To find out how many moles of salt are contained in 300. ml of a 0.40 M NaCl solution, we start with the volume in liters (0.300 L) and multiply it by the number of moles per liter of solution, which is 0.40 moles over 1.00 L. The answer is 0.12 moles of NaCl. This could also have been done using algebra by writing down the general equation relating molarity, moles and liters, substituting the known values, and then solving the equation for moles.

Molarity Calculations: Practice a. How would you prepare 100. mL of 0.25 M KNO3 solution? b. A chemist dissolves 98.4 g of FeSO4 in enough water to make L of solution. What is the molarity of the solution? c. How many moles of KBr are in 25.0 mL of a 1.23 M KBr solution? d. Battery acid is generally 3 M H2SO4. Roughly how many grams of H2SO4 are in 400. mL of this solution?

Dilution  Quite often, however, solutions are prepared by diluting a more concentrated solution. For example, if you needed a one molar solution you could start with a six molar solution and dilute it. Consequently, you also need to be familiar with the calculations that are associated with dilutions.  Calculating new concentrations?  Calculating initial volume? moles con =moles dil M con x vol con =M dil x vol dil (M conc ) (V conc )=(M dil ) (V dil )

Calculating New Concentration: A chemist starts with 50.0 mL of a 0.40 M NaCl solution and dilutes it to mL. What is the concentration of NaCl in the new solution? (M con ) (V con ) = (M dil ) (V dil ) (0.40 M) (50.0 mL) = (M dil ) (1000. mL) (0.40 M) (50.0 mL) (1000. mL) = (M dil ) M = M dil In this example you are asked what the concentration of a solution would be if it were made by diluting 50.0 ml of 0.40 M NaCl solution to ml. As a general procedure, I recommend that you first write down the equation that is given in the first line: the molarity of the concentrated solution times the volume of the concentrated solution is equal to the molarity of the dilute solution times the volume of the dilute solution. Next, substitute the known values into the equation as shown in the next line. Then rearrange the equation to solve for the unknown value. In this case that is done by dividing both sides of the equation by ml. Last, carry out the necessary calculations to get, in this case,.020 M.

Calculating Initial Volume: A chemist wants to make 500. mL of M HCl by diluting a 6.0 M HCl solution. How much of that solution should be used? (M con ) (V con ) = (M dil ) (V dil ) (V con ) = (M dil ) (V dil ) (M con ) (V con ) = (0.050 M)(500.mL) (6.0 M) V con = 4.2 mL The question in this example is different only in that you are asked to determine a volume instead of a concentration. The process used to answer the question is the same as in the previous example. Write down the algebraic equation that represents the relationship. Rearrange the equation to solve for the unknown quantity. Substitute numbers (and units) for the known values. (The second and third steps can be reversed if you wish.) Calculate the unknown value (which now becomes known).

Dilution Calculations: Practice A How much 2.0 M NaCl solution would you need to make 250 mL of 0.15 M NaCl solution? B What would be the concentration of a solution made by diluting 45.0 mL of 4.2 M KOH to 250 mL? C What would be the concentration of a solution made by adding 250 mL of water to 45.0 mL of 4.2 M KOH? D How much 0.20 M glucose solution can be made from 50. mL of 0.50 M glucose solution?

Serial dilution

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