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

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

Then,

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

Therefore,

We perform a material balance wrt vinegar

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

$336=0.16V+432-0.36V$

$0.36V-0.16V=432-336$

$0.20V=96$

$V=96/0.20=480L$

And,

$1200-480=720L$