A chemist has on hand a supply of a 50% methyl alchohol solution and a 20% methyl alcohol solution. How much of each kind should he mix to produce 300 grams of a 30% methyl alcohol solution?
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Solution problems are "chemistry-like" problems. They are frequently stated using the terms solution and concentration (or strength), and may refer to concentration as a percent (%) solution (of the principal ingredient).
To make these concepts concrete, consider the following simple procedure: we mix 8 oz. of water with 2 oz. of salt to obtain a solution of salt in water (commonly called "saltwater" or "brine"):
In this picture (as in all to follow), we show the constituents of the mixture as being "separated" so we can see their numerical relationshops. Of course, in a real brine solution, the salt and water would be completely mixed and indistinguishable one from the other.
The salt is referred to as the solute, the water is the solvent, and the resulting mixture as the solution. We need a way of quantifying the concentration of salt (that is, a number that expresses how salty the salt water really is). "Concentration" is defined to be
the ratio of the amount of solute (salt), to the total amount of solution (water + salt)
Note that concentration is usually expressed as a percent (obtained from the decimal concentration by multiplying by 100). In general, in the statement of problems and their answers, concentrations will be expressed using %. But in any calculations involving concentrations, we will always use the decimal form of the concentration.
In the next example, we are presented with 100 grams of a 20% solution of salt in water:
It is reasonable to ask the question "How much salt is there in this solution?". Mathematically, we are asking: "What is 20% of 100?". With the reminder that "of" usually is translated into the mathematical operation of "times" (multiplication), we then refine the question to "What is 20% times 100?". Finally, by converting the 20% to decimal form, we can answer the original question:
amount of salt = (.20)(100) = 20 grams
We can compute the amount of salt (solute) in the solution by multiplying the concentration (converted to decimal form by dividing by 100) by the total amount of solution.
Note that the solute in any solution problem can be identified as that component of the solution referred to using the % sign; for example, if a problem refers to a 7% solution of pure hydrochloric acid in water, the hydrochloric acid is the solute, and the water is the solvent.
Concentration will always be a number on the range from 0% - 100% (0.0 - 1.0 in decimal form).
A concentration of 0% (0.0) represents the absence of any solute. We can, if we wish, refer to pure water as a 0% concentration of salt in water!
On the other hand, a 100% solution represents pure solute (no solvent). A 100% solution of ethyl alcohol is pure (undiluted) ethyl alcohol.
Our calculational formulas still hold in these extreme cases: if we have 13 oz. of a 100% solution of anti-freeze, the amount of anti-freeze is given by (1.0)(13) = 13 oz. of anti-freeze (naturally!). On the other hand, 13 oz. of a 0% solution of anti-freeze contains (0.0)(13) = 0 oz. of anti-freeze (naturally!).
The resulting number is usually expressed as a % by multiplying by 100.
According to the above discussion, we measure concentration by the following formula:
(concentration) =
By rearranging this equation, we have
(amount of solute) = (concentration)(total amount of solution)
It was this form of the math model that was used in the opening discussion to compute the amount of salt in 100 grams of a 20% sulution of salt in water:
This is the math model that will be used extensively in what follows. Concentration as used in this formula (and in any equation) will be the decimal form of concentration (a number on the range 0.0 - 1.0).
So far, we have discussed situations involving only one solution at a time. Most solution word problems create a little more interest by referring to two (or more) solutions of different concentrations that are combined to form a new solution of yet another concentration.
For example, consider the following scenario:
10 oz. of a 20% solution of salt in water (solution A) is combined with 12 oz of a 50% solution of salt in water (solution B) to create a combined solution (solution C).
It is not too hard to figure out the amount of salt in solution C:
(amount of salt in solution C) = (amount of salt in solution A) + (amount of salt in solution B) =
(.20)(10) + (.50)(12) = 2 + 6 = 8 oz. of salt
In fact, we can also compute the concentration of salt in solution C. Since the total amount of solution C is 10 + 12 = 22 oz., of which 8 oz. is salt (we just computed that), we have:
(concentration of salt in solution C) = 8/22 = .36 (36%)
Summing up, we can state that the combined solution (solution C) consists of 22 oz. of a 36% solution of salt in water.
Consider our opening problem:
A chemist has on hand a supply of a 50% methyl alchohol solution and 20% methyl alcohol solution. How much of each kind should he mix to produce 300 grams of 30% methyl alcohol soluton?
There is a fundamental property possesed by this problem situation (and all others like it). We have already implicitly used this property in the above discussion "Combining Solutions". Note well: the first component of the solution has a certain amount of alcohol, the second component has a certain amount of alcohol, and we are combining these two solutions. It stands to reason that, since this process neither creates or destroys alcohol,
Solute Equation
The corresponding rough equation for our opening problem will be:
(amt. alcohol in the 50% sol.) + (amt. alcohol in the 20% sol.) = (amt. alcohol in the 30% sol.)
In the above exercises, you were able to use the math model to compute concentration (or, in its other form, to compute the amount of solute) in situations where only numerical quantities were involved. In the solution of algebra problems, the various elements of the problem are not always expressible numerically, because of the presence of an "unknown". Consider again our opening problem:
The final solution of this problem will start with a statement that identifies one of the unknowns of the problem, for example:
let x = number of oz. of 50% alcohol solution to be used
Accordingly, in the course of solving the problem, x can be used, wherever it is needed, to represent the amount of 50% alcohol to be used. But how much pure alcohol is in this x ounces of 50% alcohol? Of course, if we knew that x = 100, we could readily compute the amount of pure alcohol: (.50)(100) = 50 oz. of alcohol. But we don't know what x is (after all, it is the unknown!). The best we can do is multiply .50 times the x and express the result as an algebraic expression, not a number. So what we get is
(.50)(x) = .50x (oz. of pure alcohol in x ounces of 50% alcohol solution)
This may not seem very informative, but it is:
(1) correct (by virtue of the math model) (2) the best we can do for now (3) very handy in finding out later exactly what x is!
Unknowns can be combined in other ways to create "correct" algebraic expressions to be used in our "calculations".
Suppose we mix 10 oz. of solution A with a certain amount of solution B to obtain 30 oz. of the combined solution. How much solution B was used? I think we would all agree that the answer is 20 oz. How did you get that answer? You probably did it very quickly, almost without thinking about it, as follows:
(amt. of solution B) = (combined amt.) - (amt. of solution A) = 30 - 10 = 20 oz.
On the other hand, suppose we don't know the actual amount of solution A -- only that it is an unknown which we have called x. How do we now answer the above question?
This is correct, and the best we can do.
Returning again to our opening problem:
As before, we start with:
let x = amount of 50% alcohol to be used
300 - x (grams)
With the above preparation, we are ready to attack our opening problem, and see it through to completion:
(1) the amount of 50% alcohol, and (2) the amount of 20% alcohol to be combined to form the 30% solution.
#1: let x = the amount of 50% alcohol to be used
#2: N/A
#3: rough equation: (amount of alcohol in 50% solution) + (amount of alcohol in 20% solution)
This will turn our rough equation into the following:
(conc. of 50% solution)(amt. of 50% solution) + (conc. of 20% solution)(amt. of 20% solution)
(1) The various concentrations are easy: .50, .20, and .30 respectively. (2) The amount of 50% solution is also easy: it is simply x (see the "let statement"!). (3) But what about the amount of 20% solution? Since the total amount is to be 300 ozs., that leaves 300 - x for the amount of 20% solution.
finished equation:
#4:
.50x + (.20)(300) - .20x = 90 .30x = 90 - 60 .30x = 30 x = 100 check: .50(100) + (.20)(300 - 100) ?= .30(300) 50 + 40 ?= 90 (it checks)
Since the question was "how much of each solution must we mix", we must provide two "answers" . The answer to the amount of 50% solution is given by the solution to our equation: x = 100 oz.. What about the amount of 20% solution? Since there is to be 300 oz. altogether, this leaves 200 oz. for the 20% solution. Our answer looks like this:
#5: 100 oz. of 50% methyl alcohol, 200 oz. of 30% methyl alcohol.
Note that we can check our answer by doing the following calculations:
amt. alcohol in 50% solution: 50% of 100 oz. = 50 oz of alcohol amt. alcohol in 20% solution: 20% of 200 oz. = 40 oz. of alcohol for a total of 90 oz of alcohol in the resulting mixture.
90/300 = .30 (30%)
We repeat the whole solution as it should be written out by the solver: