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A laboratory employee is mixing a 10% saline solution with 4% saline solution. How much of each solution is needed to make 500 milliliters of a 6% solution. 170 milliliters of the 10% solution and 340 milliliters of 4% solution 167 milliliters of the 10% solution and 333 milliliters of 4% solution 110 milliliters of the 10% solution and 210 milliliters of 4% solution 155 milliliters of the 10% solution and 300 milliliters of 4% solution

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irspow
let t be 10% solution and f be 4% solution...

(0.1t+0.04f)/500=0.06  and t+f=500 so f=500-t making our equation:

(0.1t+0.04(500-t))=30

0.1t+20-.04t=30

0.06t=10

t=166 2/3 ml and since f=500-t, f=333 1/3 ml

So they obviously rounded to nearest ml...

167ml of 10% and 333ml of 4%

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