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soniadiana2011
- Junior | Next Rank: 30 Posts
- Posts: 14
- Joined: Wed May 16, 2012 8:52 am
Can someone please explain me the Mixture Problems on page 61 and page 68 of Flashcards?
I'm glad I can help you with this and all future mixture problems. I'd encourage you to go back and take a look at those problems once I tell you the strategy and explain the solution.
How many liters of a solution that is 15% salt
must be added to 5 liters of a solution that is 8% salt so
that the resulting mixture is 10% salt?
So, going forward, always remember that in a mixture, the components will remain constant. In my chemical engineering/ chemistry language, a constant quantity will remain constant no matter what.
Here the salt content will remain constant. For example, If I added 10 liters of sugar syrup in 5 liters of sugar syrup, I would have 15 liters of sugar syrup solution, no matter how much water I add!
So, in the question above, I know that there is a solution with 15% salt => In a 100 liter solution, 15 liters will be salt (assuming its liquid
). So 15 is constant! So, to make things simpler, I'll get a constant quantity in units our human minds can understand -> A Per Liter value!15/100=> 0.15 (liter salt)/(liter solution) => 0.15 salt per liter is constant
8% salt in second solution means = In a 100 liter solution, 8 liters will be salt. So, 8 is constant too. No matter what i do, every 100 liters will contain 8 liters of salt. Again, lets find a per liter value = 8/100 = 0.08 (liter salt)/(liter solution). 0.08 salt per liter is constant.
What the two per liter solution values tell me is that, Now if i had 10 liters of the second solution (the one with 8% salt), I can find out how much actual salt there is in that 10 liters => 0.08*10 = 0.8 liters or grams or whatever unit.
Here it is GIVEN that 5 liter of the second solution exists => how much salt do we already have in this solution? 0.08*5 = 0.4 liters!
And we want to know how much of the first solution should i mix, so the resulting mixture is 10% salt! Ok.
What I am really saying is that, i want to mix 'n' liters of the first solution to 5 liters of the second solution. So how much would the total solution in liters be? (5+n) liters !
And out of that (5+n) liters 10% should be salt => 10/100=0.1 (liter salt)/(liter solution) should be there. K so that means (5+n) liters of solution will have how much salt ? 0.1*(5+n) liters of salt.
Now we use mathematics. Remember i told you the salt will remain constant. For example, if i add 5g of sugar in a syrup that had 10 grams of sugar, the total will be 15 grams! Right? So lets make equations:
Salt from solution 1 + Salt from Solution 2 = Total salt in the new solution
0.15n + 0.08(5) = 0.1(n + 5)
15n + 40 = 10n + 50
5n = 10 => n = 2 liters
K. So its almost 6 in the morning here and I really have to go to sleep. Please try the other question yourself and do inform me if you still don't understand how to do these kinds of questions. In the question on page 68, balance the alcohol content! Good luck. Hope this helped.
