Word problems involving mixtures
By DarthVader
Date: 2021-10-06
Topic: 43 see comments
Post views: 1235
For problems involving mixtures there is an equation: c¹v¹ + c²v² = c³v³ 👌 (c) = Concentration (v) = Volume For example: [Jessica mixed 12 mL of an acidic solution with 9 mL of a 42% acidic solution, to make a 45% acidic solution. Find the percent acid concentration of the first solution.] c¹v¹ is the first solution (Although c¹ is the unknown value we are trying to find, so it is left as c¹ for now) and c²v² is the second solution, while c³v³ is the resulting mixture. Substituting these values into the equation we get: c¹(12) + (42%)(9) = (45%)v³ v³ is equal to the sum of v¹ + v², which is 12 + 9 which is equal to 21. So, substituting this into v³ we get: c¹(12) + (42%)(9) = (45%)(21) Now we can use algebra to create an equation and solve for c¹, so we can substitute c¹ with an x, and perform the multiplications like so: 12x + 378 = 945 Now to isolate the variable on one side by subtracting 378 from both sides: 12x = 567 Finally divide both sides by 12: x = 47.25 So the answer is that the first solution must be a 47.25% acid solution.
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