prasun84 wrote:

\(X\) grams of water were added to 80 grams of a strong solution of acid. If as a result, the concentration of acid in the solution became \(\frac{1}{Y}\) times of the initial concentration, what was the concentration of acid in the original solution?

(1) \(X = 80\)

(2) \(Y = 2\)

Responding to a pm:

**Quote:**

I made this equation: A1/(80+x) = 1/Y *(A2/80).

Is this equation right? Where A1 is the new concentration of acid, and A2 is the original one. I realize my initial wrror , which was to put both A1 and A2 as A (lack of concentration I suppose).

I don't know how you get this equation.

Amount of acid remains same after you add water so

Amount of acid = C1*V1 (initial) = C2*V2 (final)

C1*80 = C1/Y * (80+x)

80 = (80+x)/Y

The equation is independent of C1.

Look at the statements now:

(1) \(X = 80\)

If X = 80, Y = 2

But you don't get the value of C1.

(2) \(Y = 2\)

If Y = 2, you get x = 80

But you don't get the value of C1.

Answer (E)

For more details, check here:

http://www.veritasprep.com/blog/2012/01 ... -mixtures/
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