Discover answers to your most pressing questions at Westonci.ca, the ultimate Q&A platform that connects you with expert solutions. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.
Sagot :
Answer:
300 mL
Step-by-step explanation:
There is a simple diagrammatic way to solve mixture problems, illustrated in the attachment. In the left column are the available solution percentages. In the center is the percentage of the desired mix. In the column to the right of that are the diagonal differences between the desired value and the constituent values. Those differences represent the proportions of the constituents on the same horizontal line.
__
Here, that means 30 parts of 60% solution are needed for 5 parts of 25% solution. Since we have 50 mL of 25% solution, we can multiply these ratio values by 10 mL to find the quantities needed. That has been done here in the right-most column.
300 mL of 60% solution are needed to obtain the desired mixture.
__
Additional comment
If you want to write an equation, you can let x represent the quantity of high-concentration solution needed. Then the amount of alcohol in the mix is ...
0.60x +0.25(50) = 0.55(x +50)
0.05x = 0.30(50) . . . . . . . subtract 0.55x+0.25(50)
x = 300 . . . . . . . . . . . . mL of 60% solution needed
We hope this information was helpful. Feel free to return anytime for more answers to your questions and concerns. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.