word problem

Question

A chemist needs 5 liters of a 20% alcohol solution but has only a 25% alcohol solution. How many liters each of the 25% solution and water sh0uld mix to make the desired 5 liters of 20% solution?

Answer 1

That is a good logical approach to the right answer.

 

A logical or mathematical solution may appeal to different students.

A more mathematical equation type of solution could be:

 

Make a Table using the equation:

Amount of Mixture X Mixture Percentage = Quantity of Substance

1) 5 liters – W 25 % 0.25 x ( 5 liters – W )

2) W 0 % + 0 x W = 0

Total 5 liters 20 % 0.20 x ( 5 liters)

0.25 x ( 5 liters – W ) + 0 = 0.20 x ( 5 liters)

do some algebra

W = [ 0.20 x ( 5 liters) – 0.25 ( 5 liters ) ] / ( – 0.25 ) = 1 liter

So 1 liter of water and 4 liters of 25% solution

OR a ratio based way:

X = % amount of 25% alcohol (-OH)

 

25% is 25 parts out of 100 parts

20% is 20 parts out of 100 parts

 

X  •   25 parts -OH  =   20 parts -OH

       100 parts            100 parts

 

X = 20   •   100   =   20   =  0.80  =  80%

      100     25      25

 

SO the new solution needs to be 80 % of 25% -OH

so mix 80 parts of 25% -OH with 20 parts of H20

 

80% of 5 L is 4 L and 20% of 5 L is 1 L

so mix 4L 25% -OH with 1 L H2O

Answer 2

Hi Cecil,

The alcohol content of 5 L of 20% alcohol is 1 L of "pure" alcohol.

To get that amount of "pure" alcohol from the 25% alcohol solution, one needs 4L.

So to make a total of 5L, only 1 L of water needs to be added to the 4 L of 25% alcohol.