A chemist wants to mix a 65% alcohol solution with 10 liters of a 45% alcohol solution to produce a solution that is 55% alcohol. How many liters of the 65% alcohol solution should be used?

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Answer 1 · 2018-07-19T12:28:08

#10# liters of #65%# alcohol solution to be added.

Let #x# liters of #65%# alcohol solution is added to #10# liters of

#45%# alcohol solution to get resulting solution of #55%# alcohol.

Balancing alcohol content we get,

# x * 65/100+ 10 * 45/100 =( x+10)* 55/100# or

#65 x +450 = 55(x+10) or 65 x- 55 x= 550 -450# or

#10 x =100 or x=10# liters

Hence, #10# liters of #65%# alcohol solution to be added [Ans]

Answer 2 · 2018-07-19T12:36:56

#10# liters

Let #x# liters of #65%# (by volume) alcohol solution be mixed with #10# liters of #45%# (by volume) alcohol solution then

The amount of alcohol in the mixture

#=\text{65 % of x}+\text{45% of 10 litres}#

#=65/100\times x+45/100\times 10#

#=0.65 x+4.5\ \text{litres}#

The amount of water in the mixture

#=\text{(100 - 65) % of x}+\text{((100 - 45) % of 10 litres}#

#=35/100\times x+55/100\times 10#

#=0.35 x+5.5\ \text{litres}#

Now, the percentage #(%)# amount of alcohol in the mixture

#=\frac{\text{volume of alcohol in the mixute}}{\text{total volume of mixure}}#

#=\frac{0.65x+4.5}{x+10}\times 100#

But it is given that mixture has #55%# alcohol hence we have

#\frac{0.65x+4.5}{x+10}\times 100=55#

#0.65x+4.5=0.55x+5.5#

#0.1x=1#

#x=1/0.1#

#x=10#

hence #x=10# liters of #65%# alcohol solution is required