Question

105 mL of pure water (4°C) is saturated with NH, gas, producing a solution of
density 0.9 g/mL. If this solution contains 30% of NH, by weight, calculate its
volume.



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Answer 1

Answer:

Let V (mL) be the unknown volume of density 0.9 g/mL.

Density = mass / volume.

Therefore, mass of solution = 0.9V g

Given 105 mL of water of density 1 g/mL, its mass = 1 (g/mL) * 105 (mL) = 105 g. But mass of water is also equal to 70% of mass of solution, i.e. (70/100) * 0.9V g.

Equating masses of water : 105 = (70/100) * 0.9V Therefore, V = 105 * 100 / (0.9 * 70) = 166.7 mL.

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Answer 2

Answer:

166.7 ml

Explanation:

Let V (mL) be the unknown volume of density 0.9 g/mL.

Density = mass / volume.

Therefore,

Mass of solution = 0.9V g.

Given,

105 mL of water of density 1 g/mL,

Its mass = 1 (g/mL) * 105 (mL) = 105 g.

But mass of water is also equal to 70% of mass of solution, i.e.

(70/100) * 0.9V g.

Equating masses of water : 105 = (70/100) * 0.9V

Therefore,

V = 105 * 100 / (0.9 * 70) = 166.7 mL

Here we got answer....