X grams of water were added to the 80 grams of a strong solution of acid. As a result, the concentration of acid in the solution became 1/Y times of the initial concentration. What was the concentration of acid in the original solution?
1. X = 80
2. Y = 2
mixed %
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Ans is E
based on given condition we have -
let c1 be concentration
wt. of solute in the mixture = 80c1/100 gm
hence the new concentration c2 after adding X gm of water
((80c1/100) * 100) / (80+x) = c2 = c1/Y
hence we get,
80y = 80+x
From 1, we get x = 80 and y =2
but we still cant find the concentration
From 2, same as above
Hence E
based on given condition we have -
let c1 be concentration
wt. of solute in the mixture = 80c1/100 gm
hence the new concentration c2 after adding X gm of water
((80c1/100) * 100) / (80+x) = c2 = c1/Y
hence we get,
80y = 80+x
From 1, we get x = 80 and y =2
but we still cant find the concentration
From 2, same as above
Hence E
I think the OA is C. Can somebody please verify.
This is how -
Let the solution contain Z gms of Acid.
Thus concentration is Z/80
X gms water added, thus total solution = 80 + X
Thus concentration is Z/(80+X)
Z/(80+X) = (1/Y)(Z/80)
1) X = 80. Insufficient
2) Y = 2. Insufficient
Replacing both X and Y gives us the vlaue of Z. And thus can find Z/80
Answer is C
This is how -
Let the solution contain Z gms of Acid.
Thus concentration is Z/80
X gms water added, thus total solution = 80 + X
Thus concentration is Z/(80+X)
Z/(80+X) = (1/Y)(Z/80)
1) X = 80. Insufficient
2) Y = 2. Insufficient
Replacing both X and Y gives us the vlaue of Z. And thus can find Z/80
Answer is C