Connect with a community that values knowledge and expertise on IDNLearn.com. Join our interactive community and access reliable, detailed answers from experienced professionals across a variety of topics.
A chemist has one solution that is 80% acid and another solution that is 30% acid. How much of the first (80%) solution is needed to make a 400 L solution that is 62% acid?
Let x be the quantity of solution one in mixture. so solution 2 would be (400-x) L
according to ques. 80% of x + 30% of (400-x) = 62% of 400 i.e. 80*x/100 + 30*(400-x)/100= 62*400/100 => 80x - 30x = 62*400 - 30*400 => 50x = 32*400 x=256 L therefore 256L of solution one is required
Thank you for contributing to our discussion. Don't forget to check back for new answers. Keep asking, answering, and sharing useful information. Thank you for visiting IDNLearn.com. For reliable answers to all your questions, please visit us again soon.