What is the molality of a solution made by mixing 15 grams of Ba(OH)2 in 250 grams of water?

The answer to your question is molality = 0.35

Explanation:

Data

Molality = ?

mass of Ba(OH)₂ = 15 g

mass of water = 250 g

Process

1.- Calculate the molar mass of Ba(OH)₂

Ba(OH)₂ = 137 + 32 + 2 = 171 g

2.- Convert the mass to moles

                171 g 1 mol

                  15 g   x

                   x = (15 x 1) / 171

                   x = 15/171

                   x = 0.088 moles of Ba(OH)₂

3.- Calculate molality

Formula

molality = moles/mass of solvent (kg)

Substitution

molality = 0.088 / 0.250

Result

molality = 0.35

if the concentration of products is decreased, the equilibrium will shift toward the products). add an inert gas (one that is not involved in the reaction) to the constant-volume reaction mixture: this will increase the total pressure of the system, but will have no effect on the equilibrium condition.

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