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September 15th, 2008, 05:51 PM  #1 
Newbie Joined: Sep 2008 Posts: 1 Thanks: 0  Mixture Problem, please help
Question is: Brine is a solution of salt and water. If a tub contains 50 lbs of a 5% solution of brine, how much water must evaporate to change it to an 8% solution? Thank you in advance for your help in showing me step by step. Horizont 
September 16th, 2008, 05:12 AM  #2 
Senior Member Joined: Jul 2008 Posts: 895 Thanks: 0  Re: Mixture Problem, please help
How much salt would be in 50lb of a 5% solution? How much water? With that much salt, how much water do you need for an 8% solution? Calculate the difference. 
September 16th, 2008, 05:15 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 21,119 Thanks: 2331 
Original mass  new mass = 50lb  (5%/8%)50lb = 18¾lb.

September 16th, 2008, 11:12 AM  #4 
Senior Member Joined: Jul 2008 Posts: 895 Thanks: 0  Re: Mixture Problem, please help
This is interesting. I was just in conversation in another forum about the merits of directing someone to a solution vs simply doing it for them. OK, I give up, It's like trying to get someone to quit smoking. They have good reasons [to them] for not doing so. That's their choice, I suppose.


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