Chemistry

Question

It is necessary to have a 40% antifreeze solution in the radiator of a certain car. the radiator now has 70 liters of 20% solution. how many liters of this should be drained and replaced with 100% antifreeze to get the desired strength?

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2 Answer

  • The answer is : 17.5 liters drained and replaced by 17.5 liters of 100% solution. 
    x = amount drained and replaced
    (70-x) = remaining amount of 20% solution
    .20(70-x) + 1.00(x) = .40(70)
    14 - .2x + 1x = 28
    1x - .2x = 28 - 14
    .8x = 14
    x = 14/.8
    x= 17,5 ( 17.5 liters drained and replaced by 17.5 liters of 100% solution)

  • Answer:

    17.5 litres removed and 17.5 litres of pure antifreeze added

    Explanation:

    Let k equal the amount of the solution to be removed

    amount of antifreeze to be added

    0.2(70 - x) + x = 0.4(70)

    14 - 0.2x +x = 28

    0.8x = 28 -14

    x = 14/0.8 = 17.5

    x = 17.5 litres removed and 17.5 litres of pure antifreeze added