QUESTION IMAGE
Question
- kyle is doing a chemistry experiment and has 30 ml of a 25% solution. how many ml of a 80% solution should he add to obtain a mixture that is a 50% solution?
Step1: Calculate the amount of solute in the initial solution
The initial solution is 30 ml and 25% of it is solute. So the amount of solute in the initial solution is $30\times0.25 = 7.5$ ml. Let the volume of the 80% solution to be added be $x$ ml. The amount of solute in the 80% solution is $0.8x$ ml.
Step2: Set up an equation based on the concentration - mixture formula
The total volume of the final mixture is $(30 + x)$ ml, and the desired concentration of the final mixture is 50%. So the amount of solute in the final mixture is $0.5(30 + x)$ ml. According to the conservation of the amount of solute, we have the equation $7.5+0.8x=0.5(30 + x)$.
Step3: Expand and solve the equation
Expand the right - hand side: $7.5+0.8x = 15+0.5x$.
Subtract $0.5x$ from both sides: $7.5 + 0.8x-0.5x=15+0.5x - 0.5x$, which simplifies to $7.5 + 0.3x=15$.
Subtract 7.5 from both sides: $0.3x=15 - 7.5$, so $0.3x = 7.5$.
Divide both sides by 0.3: $x=\frac{7.5}{0.3}=25$ ml.
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25 ml