Learning Objectives To define a solution precisely, we need to state its concentration: how much solute is dissolved in a certain amount of solvent. Words such as dilute or concentrated are used to describe solutions that have a little or a lot of dissolved solute, respectively, but these are relative terms whose meanings depend on various factors.
Concentration is the measure of how much of a given substance is mixed with another substance. Solutions can be said to be dilute or concentrated. When we say that vinegar is \(5\%\) acetic acid in water, we are giving the concentration. If we said the mixture was \(10\%\) acetic acid, this would be more concentrated than the vinegar solution.  A concentrated solution is one in which there is a large amount of solute in a given amount of solvent. A dilute solution is one in which there is a small amount of solute in a given amount of solvent. A dilute solution is a concentrated solution that has been, in essence, watered down. Think of the frozen juice containers you buy in the grocery store. What you have to do is take the frozen juice from inside these containers and usually empty it into 3 or 4 times the container size full of water to mix with the juice concentrate and make your container of juice. Therefore, you are diluting the concentrated juice. When we talk about solute and solvent, the concentrated solution has a lot of solute versus the dilute solution that would have a smaller amount of solute. The terms "concentrated" and "dilute" provide qualitative methods of describing concentration. Although qualitative observations are necessary and have their place in every part of science, including chemistry, we have seen throughout our study of science that there is a definite need for quantitative measurements in science. This is particularly true in solution chemistry. In this section, we will explore some quantitative methods of expressing solution concentration.
There are several ways of expressing the concentration of a solution by using a percentage. The mass/mass percent (% m/m) is defined as the mass of a solute divided by the mass of a solution times 100: \[\mathrm{\% \:m/m = \dfrac{mass\: of\: solute}{mass\: of\: solution}\times100\%}\] mass of solution = mass of solute + mass solvent If you can measure the masses of the solute and the solution, determining the mass/mass percent is easy. Each mass must be expressed in the same units to determine the proper concentration. Suppose that a solution was prepared by dissolving \(25.0 \: \text{g}\) of sugar into \(100.0 \: \text{g}\) of water. The mass of the solution is mass of solution = 25.0g sugar + 100.0g water = 125.0 g The percent by mass would be calculated by: \[\text{Percent by mass} = \frac{25.0 \: \text{g sugar}}{125.0 \: \text{g solution}} \times 100\% = 20.0\% \: \text{sugar}\]
Example \(\PageIndex{1}\) A saline solution with a mass of 355 g has 36.5 g of NaCl dissolved in it. What is the mass/mass percent concentration of the solution? Solution We can substitute the quantities given in the equation for mass/mass percent: \(\mathrm{\%\: m/m=\dfrac{36.5\: g}{355\: g}\times100\%=10.3\%}\)
Exercise \(\PageIndex{1}\) A dextrose (also called D-glucose, C6H12O6) solution with a mass of 2.00 × 102 g has 15.8 g of dextrose dissolved in it. What is the mass/mass percent concentration of the solution? Answer7.90 %
Sometimes you may want to make up a particular mass of solution of a given percent by mass and need to calculate what mass of the solute to use. Using mass percent as a conversion can be useful in this type of problem. The mass percent can be expressed as a conversion factor in the form \(\frac{g \; \rm{solute}}{100 \; \rm{g solution}}\) or \(\frac{100 \; \rm g solution}{g\; \rm{solute}}\) For example, if you need to make \(3000.0 \: \text{g}\) of a \(5.00\%\) solution of sodium chloride, the mass of solute needs to be determined. Solution Given: 3000.0 g NaCl solution 5.00% NaCl solution Find: mass of solute = ? g NaCl Other known quantities: 5.00 g NaCl is to 100 g solution The appropriate conversion factor (based on the given mass percent ) can be used follows: To solve for the mass of NaCl, the given mass of solution is multiplied by the conversion factor. \[g NaCl = 3,000.0 \cancel{g \: NaCl \:solution} \times \frac{5.00 \:g \: NaCl}{100\cancel{g \: NaCl \: solution}} = 150.0g \: NaCl\] You would need to weigh out \(150 \: \text{g}\) of \(\ce{NaCl}\) and add it to \(2850 \: \text{g}\) of water. Notice that it was necessary to subtract the mass of the \(\ce{NaCl}\) \(\left( 150 \: \text{g} \right)\) from the mass of solution \(\left( 3000 \: \text{g} \right)\) to calculate the mass of the water that would need to be added.
Exercise \(\PageIndex{1}\) What is the amount (in g) of hydrogen peroxide (H2O2) needed to make a 6.00 kg , 3.00 % (by mass) H2O2 solution? Answer 180. g H2O2 Contributors and Attributions
Exercises (Mass/Volume Percent and g/L) 1. 5.0 grams of sugar are dissolved in 150 ml of aqueous solution. What is the m/V percent of sugar in the solution? 5/(150) * 100 % = 3.3 % . (20)/200 * 100 % = 10.0%
50 ml (50g/100 ml) = 2.5 g 4. You are using 150 mL of ether as a solvent. What mass of sulfanilamide crystals should be added to create a 10m/V% solution. x /( 150ml) = 10g/100ml y x = 15 = 15g of sulphanilamide 5. How many grams of LiOH are needed to make 25 ml of a 4.0 m/V % solution? 0.04(25) = 1.0 g 6. What mass of NaF must be mixed with 25 mL of water to create a 3. 5 m/V% solution. x/( 25) = 0.035 x = 0.875 g 7. An 800 ml solution of Kool Aid (mass 20 g) contains 780 g of water. What is the m/V percent of solute in this solution? (20g)/800 * 100% = 2.5% 8. What is the m/V percent of a solution created by adding 10 ml of olive oil (density 0.8g/ml) to 90 ml of vegetable oil? mass of olive oil = 8g/ml*10ml = 8 g 8/(10+90)* 100% = 8 % 9. If a 4000 ml solution of salt contains 40 g of salt, what is its m/V percent? 40/4000 *100% = 1.0% Concentration in g/L 1. Find the concentration in g/L for each of the following: a. 20 g of NaCl dissolved in 500 L of solution 20g/500 L = 0.04 g/L b. 2.8 g of NaBr dissolved in 200 mL of solution 2.8g/0.200L = 14g/L c. 200 mg of KCl dissolved in 75 mL of solution 0.200g/0.075L = 2.67 g/L Note that 1 mg/ml = 1 g/L 2. How many grams of Br2 are needed to make 250 mL of a 4.5 g/L solution? Mass = CV m = 4.5 g/L (0.250 L) = 1.1 g 3. How many grams of HCl are needed to make 500 mL of a 2 g/L solution? Mass = CV m = 2 g/L (0.500L) = 1.0 g 4. How many grams of LIF are needed to make 2.0 L of a 5 g/L solution? m = 10 g 5. What volume is needed to create a 2.0 g/L solution from dissolving 6.0 g of salt? m = CV 6 = 2V V = 3.0 L 6. Step by step, explain how you would actually prepare 2.0 L of a 5 g/L solution of NaCl in the lab. 1. Using a balance, weigh out 10.0 g of NaCl crystals. 2. Dissolve in less than 2.0 L in a beaker. 3. Transfer to a volumetric flask. Rinse beaker into flask. 4. Dilute to flask’s white mark with water. 5. Mix. |
What is the percent by mass of a solution if 15g of sugar is dissolved to make a180g of solution
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