Achemist has 300 grams of 20% hydrochloric acid solution. he wishes to drain some and replace it with an 80% solution so as to obtain a 25% solution. how many grams must he drain and replace with the 80% solution?

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The chemist has to drain 25 grams of the 20% hydrochloric acid solution and replace with the 80% solution

Solution:

Let us first set up a table and fill in the known values given

The table is attached below

Let "x" be the amount in grams for 20% acid solution

Let "y" be the amount in grams for 80% acid solution

From the given table, we can set up two equations

Sum of values of two acids = Value of mixture  

0.2x + 0.8y=75  

For convenience, we'll multiply the entire equation by 10,

2 x + 8 y = 750

x + 4y = 375 ------ eqn (1)  

Now, Sum of amounts of each acid = Amount of mixture  

x + y = 300 --------- eqn (2)

Subtracting equation (2) from (1),  

( + ) x + 4 y = 375

( − )  x +  y = 300

− − − − − − − −

( = ) 0 + 3y = 75  

Thus, 3y = 75

y = 25  

Substituting y = 25 in eqn (2),

x + 25  = 300

x = 300 – 25 = 275  

So, we have x = 275 and y = 25

Here "y = 25" represents amount in grams for 80% acid solution.

We can conclude that he has to drain 25 grams of the 20% acid solution.

derp.

Step-by-step explanation:

First guy wrong

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