How many liters of 11% solution can be made my mixing the two together?
2007-02-20 02:38:09 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Assuming he uses ALL of the 64.9% solution, let x = needed amount of 6.6%.
64.9(26.7) + 6.6x = 11(26.7+x)
64.9(26.7) + 6.6x = 11(26.7) + 11x
64.9(26.7) - 11(26.7) = 11x - 6.6x
53.9(26.7) = 4.4x
x = 327.075 L.
2007-02-20 02:50:06 · answer #1 · answered by Philo 7 · 0⤊ 0⤋
not given the amount of 6.6% solution so im assuming unlimited supply.
t=total solution in liters
.066(t-26.7)+.649(26.7)=.11(t) the sums of the solutes = the total amount of solute after mixed
.066t-1.7622+17.3283=.11t
.066t+15.5661=.11t
15.5661=.11t-.066t
15.5661=.044t
353.775=t
353.775 liters
2007-02-27 23:38:32 · answer #2 · answered by Catman 4 · 0⤊ 0⤋
6.6% of 26.7L is 1.7622
11% of 26.7L is 2.937
64.9% of 26.7 is 17.35
You have to convert dosages expressed in percent first.
Apply the formula: D(desired) / A(available) * Q (quantity)
1.7622 / 2.937 * 17.35 = 10.41L
I'm not so sure about this, but that's how we compute meds in the hospital.
2007-02-20 11:03:28 · answer #3 · answered by Ém 3 · 0⤊ 1⤋