A vinegar-water solution is used to wash windows. If 1200L of a 28% solution are required, but only 16% and 36% solutions are available, how much of each should be mixed?

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Answer 1 · 2017-03-01T16:25:46

Volume of $16 %$ $16%$ solution is $= 480 L$ $=480L$
Volume of $36 %$ $36%$ solution is $= 720 L$ $=720L$

Let $V =$ $V=$ volume of $16 %$ $16%$ solution

Then,

$(1200 - V)$ $(1200-V)$ is the volume of the $36 %$ $36%$ solution

Therefore,

We perform a material balance wrt vinegar

$1200 \cdot 0.28 = V \cdot 0.16 + (1200 - V) \cdot 0.36$ $1200*0.28=V*0.16+(1200-V)*0.36$

$336 = 0.16 V + 432 - 0.36 V$ $336=0.16V+432-0.36V$

$0.36 V - 0.16 V = 432 - 336$ $0.36V-0.16V=432-336$

$0.20 V = 96$ $0.20V=96$

$V = \frac{96}{0.20} = 480 L$ $V=96/0.20=480L$

And,

$1200 - 480 = 720 L$ $1200-480=720L$