Question

105ml of pure water at 4ยฐc is saturated with nh3 producing a solution of30% by mass of nh3.the total weight of solution after saturation become

Answer 1

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.

Answer 2

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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