Question

The density of two liquids, A and B, are 1000. kg/m3 and 600. kg/m3, respectively. The two liquids are mixed in a certain proportion and the density of the resulting liquid is 850 kg/m3. How much of liquid B (in grams) does 1.0 kg of the mixture contain? Assume the volume of the two liquids is additive when mixed.

Answer 1

Answer:

Mass of liquid B = 271.2 gram

Explanation:

Given:

Density of liquid A = 1000 kg/m³

Density of liquid B = 600 kg/m³

Density of mixture = 850 kg/m³

Mass of mixture = 1 kg

Assume:

Volume of liquid A = Va

Volume of liquid B = Vb

So,

Volume of mixture = Va + Vb

Mass of liquid A = 1000(Va)

Mass of liquid B = 600(Vb)

Mass of mixture = Mass of liquid A + Mass of liquid B

1 = 1000(Va) + 600(Vb)

Volume of mixture = 1 / 850

So,

(1/850) = Va + Vb

Vb = (1/850) - Va

1 = 1000(Va) + 600[(1/850) - Va]

Va = 7.25 × 10⁻⁴

Vb = (1/850) - Va

Vb = (1/850) - [7.25 × 10⁻⁴]

Vb = 4.25 × 10⁻⁴

Mass of liquid B = 600(Vb)

Mass of liquid B = 600(4.25 × 10⁻⁴)

Mass of liquid B = 271.2 gram