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6 January, 06:45

How much water needs to be mixed with three liters of

45% antifreeze to dilute the concentration to 35% antifreeze?

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Answers (1)
  1. 6 January, 06:57
    0
    The volume of water needed=0.86 liters

    Explanation:

    Step 1: Form an equation

    The equation can be expressed as follows;

    (Ac*Va) + (Wc+Vw) = Fc (Va+Vw)

    where;

    Ac=initial concentration of antifreeze

    Va=volume of antifreeze in liters

    Wc=concentration of water

    Vw=volume of the water in liters

    Fc=final concentration of the antifreeze

    This expression can be written as;

    (concentration of antifreeze*volume of antifreeze in liters) + (concentration of water*volume of the water) = final concentration of the antifreeze (volume of antifreeze in liters+volume of the water in liters)

    In our case;

    Ac=45%=45/100=0.45

    Va=3 liters

    Wc=0

    Vw=unknown

    Fc=35%=35/100=0.35

    Replacing;

    (0.45*3) + (0*Vw) = 0.35 (3+Vw)

    1.35=1.05+0.35 Vw

    0.35 Vw=1.35-1.05

    0.35 Vw=0.30

    Vw=0.3/0.35=0.86

    Vw=0.86 liters

    The volume of water needed=0.86 liters
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