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?

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