Question

25% strong ammonia solution to 6m ammonia solution preparation

Answer 1

hi there

Assuming you need 100 mL of 5% ammonia,

V1 x 25 = 100 x 5. Solving the equation, V1 = 20. Therefore you need to add 20 mL of 25% ammonia to 80 mL of water to get 100 mL of 5% ammonia.

Answer 2

Concept:

• Molality of solution
• Weight by weight concentration
• Calculating solution concentrations

Given:

• The original concentration of the ammonia solution is 25% weight by weight

Find:

• The amount of water to be added which is needed to prepare an ammonia solution of 6 molal

Solution:

25% weight by weight ammonia solution

Assume that the total mass of the ammonia solution is 100g.

Since we have 25% of ammonia solution, there is 25g of ammonia in 100g of solution and there is 75 g of water.

The molar mass of ammonia is 17g.

The number of moles of ammonia in 25% solution is = 25/17 = 1.47 mol

We require a 6 molal solution.

Molality = 6 m

Molality is the number of moles of a compound per kilogram of water.

Molality = n/mass of water in kg

Mass of water = mass of water in the original solution

Mass of water = 0.075+x

Molality = 1.47/(0.075+x) = 6

1.47 = 6(0.075+x)

1.47 = 0.45 + 6x

6x = 1.02

x = 0.17 kg = 170g

We need 170g more of water to add to 25% ammonia solution to make the solution 6 molal.

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