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A chemical company has 60 liters of a 40% HNO3 solution. How many liters of pure undiluted HNO3 must the chemist add so that the resultant solution is a 50% solution?
A. 12
B. 15
C. 20
D. 24
E. 30
OA A
A chemical supply company has 60 liters of a 40% HNO3 solution. How many liters of pure...
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60 liters of 40% HNO3 solution
The concentration of HNO3 present in the solution is 40 liters in 100 liters for 40% HNO3
The concentration of HNO3 in 60 liters is
$$\frac{60\cdot40}{100}=24\ liters$$
Let the volume of pure undiluted HNO3 = x
Volume of pure HNO3 to add in order to increase concentration =>
24 + x = 50% of (60+x)
24 + x = (1/2) (60+x)
2(24+x) = 60+x
48 + 2x = 60 + x
2x - x = 60 - 48
x = 12 liters of pure HNO3 must be added in order to have a concentration of HNO3 solution.
Answer = option A
The concentration of HNO3 present in the solution is 40 liters in 100 liters for 40% HNO3
The concentration of HNO3 in 60 liters is
$$\frac{60\cdot40}{100}=24\ liters$$
Let the volume of pure undiluted HNO3 = x
Volume of pure HNO3 to add in order to increase concentration =>
24 + x = 50% of (60+x)
24 + x = (1/2) (60+x)
2(24+x) = 60+x
48 + 2x = 60 + x
2x - x = 60 - 48
x = 12 liters of pure HNO3 must be added in order to have a concentration of HNO3 solution.
Answer = option A
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We add 60 liters of 40% solution and n liters of 100% solution, obtaining (60 + n) liters of 50% solution. We can create the following equation:
60 x 0.4 + n =0.5(60 + n)
24 + n = 30 + 0.5n
0.5n = 6
n = 12
Answer: A
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