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How much water must be added to a 100 cc of 80% solution of boric acid to reduce it to a 50% solution?
- a)20 cc
- b)40 cc
- c)80 cc
- d)60 cc
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
How much water must be added to a 100 cc of 80% solution of boric acid...
Quantity of boric acid in 100 cc solution = 80% of 100 cc = 80 cc
Quantity of water in 100 cc solution = 100 cc - 80 cc = 20 cc
Let x cc of water be added to get the concentration of 50%.
So, (80)/(100 + x) = 50/100
or (80)/(100 + x) = 1/2
or 80 × 2 = (100 + x)
or x = 160 - 100 = 60 cc
Thus, 60 cc of water must be added.
Hence, option (d) is correct.
Quantity of water in 100 cc solution = 100 cc - 80 cc = 20 cc
Let x cc of water be added to get the concentration of 50%.
So, (80)/(100 + x) = 50/100
or (80)/(100 + x) = 1/2
or 80 × 2 = (100 + x)
or x = 160 - 100 = 60 cc
Thus, 60 cc of water must be added.
Hence, option (d) is correct.
Alternate method:
Concentration of boric acid in 100 cc solution = 80% of 100 cc = 80 cc
Quantity of water in 100 cc solution = 100 cc - 80 cc = 20 cc
Here we are adding water.
So, amount of boric acid is fixed.
When water is added to the solution, boric acid reduces to 50% of the solution.
Thus, 50% of the solution contains 80 cc.
So, 100% of the solution contains 160 cc.
Amount of water to be added in the solution = 160 cc - 100 cc = 60 cc
Hence, option (d) is correct.
Concentration of boric acid in 100 cc solution = 80% of 100 cc = 80 cc
Quantity of water in 100 cc solution = 100 cc - 80 cc = 20 cc
Here we are adding water.
So, amount of boric acid is fixed.
When water is added to the solution, boric acid reduces to 50% of the solution.
Thus, 50% of the solution contains 80 cc.
So, 100% of the solution contains 160 cc.
Amount of water to be added in the solution = 160 cc - 100 cc = 60 cc
Hence, option (d) is correct.
Most Upvoted Answer
How much water must be added to a 100 cc of 80% solution of boric acid...
Given:
- Initial volume of 80% boric acid solution = 100 cc
- Desired final concentration = 50%
Calculations:
- Let x be the amount of water to be added to reduce the concentration.
- Initial amount of boric acid in the solution = 80% of 100 cc = 80 cc
- Final volume of the solution after adding water = 100 cc + x
- Final amount of boric acid in the solution after adding water = 50% of (100 + x) cc
Equation:
- Amount of boric acid before = Amount of boric acid after
- 80 cc = 50% of (100 + x)
- 80 = 0.5(100 + x)
- 80 = 50 + 0.5x
- 0.5x = 30
- x = 60 cc
Therefore, 60 cc of water must be added to 100 cc of 80% boric acid solution to reduce it to a 50% solution. Hence, the correct answer is option D.
- Initial volume of 80% boric acid solution = 100 cc
- Desired final concentration = 50%
Calculations:
- Let x be the amount of water to be added to reduce the concentration.
- Initial amount of boric acid in the solution = 80% of 100 cc = 80 cc
- Final volume of the solution after adding water = 100 cc + x
- Final amount of boric acid in the solution after adding water = 50% of (100 + x) cc
Equation:
- Amount of boric acid before = Amount of boric acid after
- 80 cc = 50% of (100 + x)
- 80 = 0.5(100 + x)
- 80 = 50 + 0.5x
- 0.5x = 30
- x = 60 cc
Therefore, 60 cc of water must be added to 100 cc of 80% boric acid solution to reduce it to a 50% solution. Hence, the correct answer is option D.
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