mixture word prob: How much water must be added...?

[sak7704]

How much water must you add to 100ml of a 50% acid/water solution to obtain a 30% solution.?

I must create an algebraic equation and solve.

 

[galactus]

Since you're adding pure water and not a concentration, let x=amount of water.

\(\displaystyle \L\\0.50(100)=0.30(100+x)\)

 

[soroban]

Hello, sak7704!

Galactus made a common error.
. . (I've done it myself many times.)

How much water must you add to 100ml of a 50% acid/water solution to obtain a 30% solution?

Since we're adding water (not acid), we must think in terms of water.

We have a solution which is 50% water.
. . It has: \(\displaystyle 50\%\,\times\,100\:=\:50\) ml of water.

We will add \(\displaystyle x\) ml of water.
. . So there is: \(\displaystyle 50\,+\,x\) ml water.

We get a new solution which is \(\displaystyle 70\%\) water.
. . We have \(\displaystyle 100\,+\,x\) ml of solution which has: \(\displaystyle 0.70(100\,+\,x)\) ml water.

There is our equation! .The two amounts of water are equal.

. . \(\displaystyle 50\,+\,x\;=\;0.70(100\,+\,x)\)

 

[skeeter]

I disagree ... galactus made no error.

(100 ml)(50%) + (x ml)(0%) = (100 + x ml)(30%)

(100)(.5) + 0 = (100 + x)(.3) ... which is the same as his equation

50 = 30 + .3x

20 = .3x

x = 200/3 ml

soroban is also correct, he just took a different tact ...

50 + x = .7(100 + x)

50 + x = 70 + .7x

.3x = 20

x = 200/3 ml of water

 

[pka]

Galactus made a common error. ???????
\(\displaystyle 50\,+\,x\;=\;0.70(100\,+\,x)\)

Those two equations are identical.

 

[soroban]

Hello, skeeter!

Absolutely right . . . Galactus equated the acid content.
. . {I've still made the errors I accused him of.)

 

[Denis]

After the water is added, the 50 ml of acid = 30% of total contents; so:
50 = .30x
x = 166 2/3
added water = 166 2/3 - 100 = 66 2/3

...same thing, different approach...