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<H2>Mixture problems</H2> This lesson presents some typical <B>Mixture problems</B> and the methodology for their solutions. The problems gathered in this lesson are of the type "<B><U>how much of the pure solute (bulk liquid, i.e. water) should be added</U></B>...", or "<B><U>how much of water should be taken off (evaporated)</U></B>...", or "<B><U>how much of the pure solvent (salt, acid etc.) should be added</U></B>...", or "<B><U>how much of the solution with the given concentration should be added</U></B>..." . The way to solve these problems is to reduce them to the linear equation with one unknown, and then to solve this equation. <BLOCKQUOTE>The <B>Mixture problems</B> of the <B>different type</B> are presented in the lesson <A HREF = http://www.algebra.com/algebra/homework/word/mixtures/More-Mixture-problems.lesson> More Mixture problems</A> in this module. The way to solve that problems is to reduce them to the linear system of two equations in two unknowns.</BLOCKQUOTE> <H3>Problem 1. Add water to the Salt solution</H3>How much water should be added to 200 milliliters of a 10% salt solution to get a 2% salt solution? (Concentrations here are mass-to-volume concentrations, measured in [g/mL] units, grams of the salt per 1 milliliter of the solution volume). <B>Solution</B> Since an initial solution concentration is 10%, it contains 0.1 [g/mL]*200 mL = 20 g (grams) of salt. Let us denote as <B>x</B> a volume of water in milliliters, which should be added to 200 mL of the solution to get a 2% solution. After adding water, the volume of the solution will be 200 mL + x mL = (200 + x) mL. The mass of salt remains the same in the new solution, i.e. 20 grams. Since the concentration after adding water should be equal to 2%, this gives us an equation {{{20/(200 + x) = 0.02}}}. Simplify this equation step by step: {{{20 = 0.02*(200 + x)}}} (after multiplication both sides by {{{200+x}}}); {{{20 = 4 + 0.02*x}}} (after brackets opening); {{{20 - 4 = 0.02*x}}} (after collecting the constant terms on the left side); {{{0.02*x = 16}}} (after reducing like terms); {{{x = 16/0.02 = 800}}}. So, {{{x=800}}} mL is the potential solution. The last step is to check the solution. Simply substitute the value of {{{x=800}}} into the very first equation: {{{20/(200 + x) = 20/(200 + 800) = 20/1000 = 0.02}}}. The check shows that the solution is correct. <B>Answer</B>. 800 mL of water should be added to 200 milliliters of the 10% salt solution to get the 2% salt solution. <H3>Problem 2. Evaporate water from the Salt solution</H3>How much water must be evaporated from 1000 milliliters of a 2% salt solution to get a 10% salt solution? (Concentrations here are mass-to-volume concentrations, measured in [g/mL] units, same as in the <B>Problem 1</B>). <B>Solution</B> Since an initial solution concentration is 2%, it contains 0.02 [g/mL]*1000 mL = 20 g (grams) of salt. Let us denote as <B>x</B> a volume of water in milliliters, which should be evaporated from 1000 mL of the solution to get a 10% solution. After evaporating water, the volume of the solution will be 1000 mL - x mL = (1000 - x) mL. The mass of salt remains the same in the new solution, i.e. 20 grams. Since the concentration after evaporating water should be equal to 10%, this gives us an equation {{{20/(1000 - x) = 0.1}}}. Simplify this equation step by step: {{{20 = 0.1*(1000 - x)}}} (after multiplication both sides by {{{1000-x}}}); {{{20 = 100 - 0.1*x}}} (after brackets opening); {{{0.1*x = 100 - 20}}} (after collecting variable term on the left side, constant terms on the right side); {{{0.1*x = 80}}} (after reducing like terms); {{{x = 80/0.1 = 800}}}. So, {{{x=800}}} mL is the potential solution. The last step is to check the solution. Simply substitute the value of {{{x=800}}} into the very first equation: {{{20/(1000 - x) = 20/(1000 - 800) = 20/200 = 0.1}}}. The check shows that the solution is correct. <B>Answer</B>. 800 mL of water should be evaporated from 1000 milliliters of the 2% salt solution to get the 10% salt solution. <H3>Problem 3. Add the salt to the Salt solution</H3>How much salt should be added to 1000 milliliters of a 2% salt solution to get a 4% salt solution? (Concentrations here are mass-to-volume concentrations, measured in [g/mL] units, same as in <B>Problem 1</B>). <B>Solution</B> Since an initial solution concentration is 2%, it contains 0.02 [g/mL]*1000 mL = 20 g (grams) of salt. Let us denote <B>x</B> a mass of salt in grams, which should be added to 1000 mL of the solution to get a 4% solution. After adding salt, the volume of the solution remains the same, equal to 1000 mL. The mass of salt after adding salt is equal to 20 g + x g = (20 + x) g (grams). Since the concentration after adding water should be equal to 4%, this gives us an equation {{{(20+x)/1000 = 0.04}}}. Simplify this equation step by step: {{{20 + x = 1000*0.04}}} (after multiplication both sides by 1000); {{{20 + x = 40}}}; {{{x = 40 - 20}}}; {{{x = 20}}}. So, {{{x=20}}} g is the potential solution. The last step is to check the solution. Simply substitute the value of {{{x=20}}} into the very first equation: {{{(20+x)/1000 = (20+20)/1000 = 40/1000 = 0.04}}}. The check shows that the solution is correct. <B>Answer</B>. 20 g of salt should be added to 1000 milliliters of the 2% salt solution to get the 4% salt solution. <H3>Problem 4. Add water to the Acid solution</H3>How much water should be added to 200 milliliters of a 10% acid solution to get a 2% acid solution? (Concentrations here are volume-to-volume concentrations, measured in [mL/mL] units, milliliter of the acid volume per 1 milliliter of the solution volume). <B>Solution</B> Since an initial solution concentration is 10%, it contains 0.1 [mL/mL]*200 mL = 20 mL (milliliters) of acid. Let us denote as <B>x</B> a volume of water in milliliters, which should be added to 200 mL of the solution to get a 2% solution. After adding water, the volume of the solution will be 200 mL + x mL = (200 + x) mL. The volume of acid remains the same in the new solution, i.e. 20 milliliters. Since the concentration after adding water should be equal to 2%, this gives us an equation {{{20/(200 + x) = 0.02}}}. This equation is the same as in <B>Problem 1</B>. Solve it yourself as an exercise. <B>Answer</B>. 800 mL of water should be added to 200 milliliters of the 10% acid solution to get the 2% acid solution. <H3>Problem 5. Add the pure acid to the Acid solution</H3>How much of the pure acid should be added to 1000 milliliters of a 2% acid solution to get a 4% acid solution? (Concentrations here are volume-to-volume concentrations, measured in [mL/mL] units, same as in <B>Problem 4</B>). <B>Solution</B> Since an initial solution concentration is 2%, it contains 0.02 [mL/mL]*1000 mL = 20 mL (milliliters) of acid. Let us denote as <B>x</B> a volume of acid in milliliters, which should be added to 1000 mL of the solution to get a 4% solution. After adding <B>x</B> mL of acid, the volume of the solution will be equal to 1000 mL + x mL = (1000+x) mL. The volume of acid after adding acid is equal to 20 mL + x mL = (20 + x) mL (milliliters). Since the concentration after adding water should be equal to 4%, this gives us an equation {{{(20+x)/(1000 + x) = 0.04}}}. Simplify this equation step by step: {{{20 + x = 0.04*(1000+x)}}} (after multiplication both sides by 1000+x); {{{20 + x = 40 + 0.04*x}}} (after brackets opening); {{{x - 0.04*x = 40 - 20}}} (after collecting variable terms on the left side, constant terms on the right side); {{{0.96*x = 20}}} (after reducing like terms); {{{x = 20/0.96 = 20.833}}}. So, {{{x=20.833}}} mL is the potential solution. The last step is to check the solution. Simply substitute the value of {{{x=20.833}}} into the very first equation: {{{(20+x)/(1000+x) = (20+20.833)/1020.833 = 40.833/1020.833 = 0.04}}}. The check shows that the solution is correct. <B>Answer</B>. 20.833 mL of the pure acid should be added to 1000 milliliters of the 2% acid solution to get the 4% acid solution. Note that the equation and the resulting value in this Problem are different from those of <B>Problem 3</B>. <H3>Problem 6. Add the Acid solution to the Acid solution</H3>How much of the 10% acid solution should be added to 1000 milliliters of a 2% acid solution to get a 4% acid solution? (Concentrations here are volume-to-volume concentrations, measured in [mL/mL] units, same as in <B>Problem 4</B>). <B>Solution</B> Since an initial solution concentration is 2%, it contains 0.02 [mL/mL]*1000 mL = 20 mL (milliliters) of acid. Let us denote as <B>x</B> a volume of the 10% acid solution in milliliters, which should be added to 1000 mL of the solution to get a 4% solution. After adding <B>x</B> mL of the 10% acid solution, the volume of the solution will be equal to 1000 mL + x mL = (1000+x) mL. The volume of acid after adding x mL of the 10% acid solution is equal to 20 mL + 0.1*x mL = (20 + 0.1*x) mL (milliliters). Since the concentration after adding water should be equal to 4%, this gives us an equation {{{(20+0.1*x)/(1000 + x) = 0.04}}}. Simplify this equation step by step: {{{20 + 0.1*x = 0.04*(1000+x)}}} (after multiplication both sides by 1000+x); {{{20 + 0.1*x = 40 + 0.04*x}}} (after brackets opening); {{{0.1*x - 0.04*x = 40 - 20}}} (after collecting variable terms on the left side, constant terms on the right side); {{{0.06*x = 20}}} (after reducing like terms); {{{x = 20/0.06 = 333.333}}}. So, {{{x=333.333}}} mL is the potential solution. Now, check the solution. Simply substitute the value of {{{x=333.333}}} into the very first equation: {{{(20+0.1*x)/(1000+x) = (20+0.1*333.333)/(1000+333.333) = 53.333/1333.333 = 0.04}}}. The check shows that the solution is correct. <B>Answer</B>. 333.333 mL of the 10% acid solution should be added to 1000 milliliters of the 2% acid solution to get the 4% acid solution. <H3>Problem 7. Mixing water and antifreeze</H3> How much pure antifreeze liquid should be added to 1 gallon of 40% antifreeze to get 60% antifreeze? (Concentrations here are volume concentrations, measured in [volume/volume] units). <B>Solution</B> First, there is 0.4 gal of pure antifreeze in 1 gallon of 40% antifreeze. Let us denote <B>x</B> a volume of pure antifreeze, which should be added to 1 gallon of 40% antifreeze to get 60% antifreeze. So, the volume of antifreeze after adding is <B>0.4 + x</B> gallons, while the total volume of liquid (water plus antifreeze) after adding is <B>1.0+x</B>. The condition of 60% volume concentration gives an equation {{{(0.4+x)/(1.0+x) = 0.6}}}, or <B>0.4+x = 0.6*(1.0+x)</B>. This is the linear equation. (Note that this is the same form equation as in the <B>Problem 5</B>). Simplify the equation step by step: {{{0.4+x = 0.6 + 0.6*x}}}, {{{x-0.6*x = 0.6-0.4}}}, {{{0.4*x = 0.2}}}, {{{x = 0.2/0.4 = 0.5}}} gallons. <B>Check</B>: {{{(0.4+0.5)/(1.0+0.5) = 0.9/1.5 = 0.6}}}. <B>Answer</B>. 0.5 gallons of pure antifreeze should be added to 1 gallon of 40% antifreeze to get 60% antifreeze. <B>Note</B>. Among other, problems gathered in this lesson show the difference between equations for dissolving the additional salt and adding the liquid solution. When the liquid solution is added, the increase of the total volume should be counted. When the additional amount of the salt is dissolved, the change of the total volume is negligible, at least in the practical range of small or moderate concentrations. My other lessons on word problems for mixtures in this site are - <A HREF=http://www.algebra.com/algebra/homework/word/mixtures/More-Mixture-problems.lesson>More Mixture problems</A> - <A HREF=http://www.algebra.com/algebra/homework/word/mixtures/Solving-typical-mixture-problems.lesson>Solving typical word problems on mixtures for solutions</A> - <A HREF=https://www.algebra.com/algebra/homework/word/mixtures/Word-problems-on-mixtures-for-antifreeze-solutions.lesson>Word problems on mixtures for antifreeze solutions</A> - <A HREF=http://www.algebra.com/algebra/homework/word/mixtures/Solving-typical-word-problems-on-mixtures-for-dry-substances-coffee-nuts.lesson>Word problems on mixtures for dry substances like coffee beans, nuts, cashew and peanuts</A> - <A HREF=http://www.algebra.com/algebra/homework/word/mixtures/Word-problems-on-mixtures-for-dry-substances-like-cookies-candies.lesson>Word problems on mixtures for dry substances like candies, dried fruits</A> - <A HREF=http://www.algebra.com/algebra/homework/word/mixtures/Word-problems-on-mixtures-for-dry-substances-like-soil-and-sand.lesson>Word problems on mixtures for dry substances like soil and sand</A> - <A HREF=https://www.algebra.com/algebra/homework/word/mixtures/Word-problems-on-mixtures-for-alloys.lesson>Word problems on mixtures for alloys</A> - <A HREF=https://www.algebra.com/algebra/homework/word/mixtures/Typical-word-problems-on-mixtures-from-the-archive.lesson>Typical word problems on mixtures from the archive</A> - <A HREF=https://www.algebra.com/algebra/homework/word/mixtures/Advanced-mixture-problems.lesson>Advanced mixture problems</A> - <A HREF=https://www.algebra.com/algebra/homework/word/mixtures/Advanced-mixture-problem-for-three-alloys.lesson>Advanced mixture problem for three alloys</A> - <A HREF=https://www.algebra.com/algebra/homework/word/mixtures/Unusual-word-problem-on-mixtures.lesson>Unusual word problem on mixtures</A> - <A HREF=https://www.algebra.com/algebra/homework/word/mixtures/Check-if-you-know-the-basics-of-mixtures-from-Science.lesson>Check if you know the basics of mixtures from Science</A> - <A HREF=https://www.algebra.com/algebra/homework/word/mixtures/Special-type-mixture-problems-on-DILUTION-adding-water.lesson>Special type mixture problems on DILUTION adding water</A> - <A HREF=https://www.algebra.com/algebra/homework/word/mixtures/Increasing-concentration-of-an-acid-solution-by-adding-pure-acid.lesson>Increasing concentration of an acid solution by adding pure acid</A> - <A HREF=https://www.algebra.com/algebra/homework/word/mixtures/Increasing-concentration-of-alcohol-in-solution-by-adding-pure-alcohol.lesson>Increasing concentration of alcohol solution by adding pure alcohol</A> - <A HREF=https://www.algebra.com/algebra/homework/word/mixtures/How-many-kilograms-of-sand-must-be-added-to-a-mixture-of-sand-and-cement.lesson>How many kilograms of sand must be added to a mixture of sand and cement</A> - <A HREF=https://www.algebra.com/algebra/homework/word/mixtures/Draining-replacing-mixture-problems.lesson>Draining-replacing mixture problems</A> - <A HREF=https://www.algebra.com/algebra/homework/word/mixtures/How-much-water-should-be-evaporated.lesson>How much water must be evaporated</A> - <A HREF=https://www.algebra.com/algebra/homework/word/mixtures/Advanced-problems-on-draining-and-replacing.lesson>Advanced problems on draining and replacing</A> - <A HREF=https://www.algebra.com/algebra/homework/word/mixtures/Using-effective-methodology-to-solve-many-steps-dilution-problems.lesson>Using effective methodology to solve many-steps dilution problems</A> - <A HREF=https://www.algebra.com/algebra/homework/word/mixtures/Entertainment-problems-on-mixtures.lesson>Entertainment problems on mixtures</A> - <A HREF=http://www.algebra.com/algebra/homework/word/mixtures/OVERVIEW-of-lesson-on-word-problems-for-mixtures.lesson>OVERVIEW of lessons on word problems for mixtures</A> Use this file/link <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A> to navigate over all topics and lessons of the online textbook ALGEBRA-I.
