Question

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?

Answer 1

Answer:

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