Alcohol solution??????????

Question

Sir Ronald E B. . .

thanks for immediately responding to my question.

however, my apologies that i'm quite not good in math. i think i'm confused as to how you got the "6" at the right side of the equation.

thanks again

Answer 1

what he said:

X liters * 60% + (15-X) liters * 40% = 15 liters x 48%

0.60X - 0.40X = 7.20 - 6.00

0.20X = 1.20

X = 6 liters of 60%

15-6 = 9 liters of 40%

===================================

he is using a modified version of this formula:

C1V1 = C2V2

it says that concentration times volume, finds the power within a solution, just like finding the moles within say... 250 ml of 6 molar HCl:  C1V1 = (6 mol/Litre)(0.250 litres) = 1.5 moles

-----------------------------

the power within the 15 litre mix of the 60% & 40% equals the power within the 15 litres of the 48%.

it was an arbitrary decision whether to choose the 60% to be "X" litres out of the 15, or to choose the 40% to be the "X" litres. he chose the 60% to be the "X" litres, which meant the 40% must be the rest (15-X)litres

60%  &  40%   =  48%

C1V1  & C2V2  = C3V3

(60%)(X) & (40%)(15-X) = (48%)(15)

(0.60)(X) & (0.40)(15-X) = (0.48)(15)

it is at this point, that the "6" of which you inquire, shows up,.. as the multiplication product of the 40%  of the 15:

(0.60X) & (6 - 0.40X) = (7.2)

(0.60X) & ( - 0.40X) = 7.2 -6

0.20X = 1.2

X = 6 litres of the 60% solution

which means that the other 9 litres must be the 40%