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25 April, 01:42

A chemist mixes two liquids A and B to form a homogeneous mixture. The densities of the liquids are 2.0514 g/mL for A and 2.6678 g/mL for B. When she drops a small object into the mixture, she finds that the object becomes suspended in the liquid; that is, it neither sinks nor floats. If the mixture is made of 41.37 percent A and 58.63 percent B by volume, what is the density of the object? Can this procedure be used in general to determine the densities of solids? What assumptions must be made in applying this method?

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  1. 25 April, 01:55
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    a) the density of the object is 1.9302 g/mL

    b) The procedure used to determine the density of a solid, is the principle of Archimeides

    c) In order to use this procedure we had to assume a total volume of the fluid mixture and a percentage of the solid submerged in this

    Explanation:

    assuming the volume of the mixture is: vAB = 1000 mL

    ⇒ vA = 413.7 mL ∧ vB = 586.3 mL

    mix (A+B) density = mass AB / volume AB

    ∴ mAB = mA + mB

    ⇒ mA = vA * dA = 413.7 mL * 2.0514 g/mL = 848.664 g A

    ⇒ mB = vB * dB = 586.3 mL * 2.6678 g/mL = 1564.13 g B

    ⇒ mAB = 2412.795 g AB

    ⇒ mix density = 2412.795 g / 1000 mL = 2.4127 g/mL AB

    Starting from the principle of Archimides to find the density of a solid suspended in a liquid, we have:

    E = dAB * g * Vs ... (1)

    ∴ Vs : submerged solid volume;

    ∴ E : fluid weight displaced by the solid

    ∴ dAB : mix density

    ∴ g : gravity

    in the sum of forces: ∑F = 0

    ⇒ E - Wsolid = 0

    ⇒ E = Wsolid

    ∴ Wsolid = msolid * g ... (2)

    (1) = (2)

    ⇒ msolid * g = dAB * g * Vs

    ⇒ msolid = dsolid * Vs ... (3)

    ∴ msolid = dsolid * Vsolid

    assuming that 80% of the solid is submerged in the mixture, we have:

    ⇒ Vs = 0.8 * Vsolid ... (4)

    (4) in (3):

    ⇒ dsolid * Vsolid = dAB * 0.8 * Vsolid

    ⇒ dsolid = dAB * o. 8

    ⇒ dsolid = 2.4127 g/mL * 0.8 = 1.9302 g/mL
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