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28 February, 07:04

A chemist has three different acid solutions. The first acid solution contains 20% acid, the second contains 35% and the third contains 80%. He wants to use all three solutions to obtain a mixture of 54 liters containing 45% acid, using 2 times as much of the 80% solution as the 35% solution. How many liters of each solution should be used?

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  1. 28 February, 07:13
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    the volume of First acid solution = 24 liters

    the volume of second acid solution = 10 liters

    the volume of third acid solution = 20 liters

    Step-by-step explanation:

    Let,

    the volume of First acid solution be 'x'

    the volume of second acid solution be 'y'

    the volume of third acid solution be 'z'

    Data provided in the question:

    Acid in first acid solution = 20% = 0.2

    Acid in Second acid solution = 35% = 0.35

    Acid in third acid solution = 80% = 0.8

    Acid in the final solution = 45% = 0.45

    Volume of the final solution = 54 liters

    According to question

    0.2x + 0.35y + 0.8z = 0.45 * 54

    or

    0.2x + 0.35y + 0.8z = 24.3 ... (1) [concentration = Percentage * volume ]

    z = 2y ... (2)

    also,

    x + y + z = 54 ... (3) [Total volume = sum of individual solutions]

    substituting 2 in 3,

    x + y + 2y = 54

    x + 3y = 54

    or

    x = 54 - 3y ... (4)

    substituting 2 and 4 in equation 1

    0.2 (54 - 3y) + 0.35y + 0.8 (2y) = 24.3

    or

    10.8 - 0.6y + 0.35y + 1.6y = 24.3

    or

    1.35y = 13.5

    or

    y = 10 liters

    substituting the value of y in 2, we get

    z = 2 (10)

    or

    z = 20 liters

    and,

    substituting the value of y in 4, we get

    x = 54 - 3 (10)

    or

    x = 24 liters

    Hence,

    the volume of First acid solution = 24 liters

    the volume of second acid solution = 10 liters

    the volume of third acid solution = 20 liters
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