2 Answers | Add Yours
Let x be the weights of pure water to be added.
Let y be the weight of the 10% solution.
x + 100 = y
Now Let us auume that the amount of salt in the pure water is 0.
Given the amount of salt in the 30% solution is equal to the amount of salt in the final saline solution at 10%.
Then we could write:
0 + 30% 100 = 10% y
Substitute y by x + 100 in the last equation and solve.
30% 100 = 10% (x + 100)
==> 30 = (x+100)/10
==> 300 = x+ 100
==> x= 200 kg.
Weight of original Saline solution = w1 = 100 kg
Concentration of the original solution = 30%
Since the quantity of solution and additional water is given in terms of weight, we assume that the concentration is in terms of weight.
Weight of salt in the original solution = 100*(30/100) = 30 kg
When we add water to the original solution the weight of salt in continues to be 30 kg.
w2 = weight of solution with when diluted to 10 percent concentration.
(Weight of salt)*100/w2 = Concentration percentage of solution
Substituting the actual weight of salt and required concentration percentage:
30*100/w2 = 10
==> w2 = 30*100/10 = 300 kg
The water to be added = (Weight of 10% solution) - (Weight of 10% solution)
= w2 - w1 = 300 - 100 = 200kg
200 kg of pure water is to be added
We’ve answered 318,988 questions. We can answer yours, too.Ask a question