Calculate the molalities of the following aqueous solutions:. (a) 1.22 M sugar (C12H22O11) solution (density of solution = 1.12 g/ml), (b) 0.87 M NaOH solution (density = 1.04 g/ml), (c) 5.24 M NaHCO3 solution (density = 1.19 g/ml).

Molality is a measure of the number of moles of solute in a solution corresponding to 1 kg or 1000 g of solvent. This contrasts with the definition of molarity which is based on a specified volume of solution

Molarity is defined as the molar concentration is a measure of the concentration of a chemical species, in particular of a solute in a solution.

=

The number of moles = 1.22

c) Molarity = 5.24 M

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The answer to your question is below

Explanation:

a)

Concentration = 1.22 M

density = 1.12 g/ml

Process

1.- Calculate the moles of sugar

Molarity = moles / 1L, then    1.22 moles / 1L

Number of moles = 1.22

2.- Calculate the mass of solution

density = mass/volume

mass = density x volume

mass = 1.12kg/L x 1 l

mass = 1.12 kg

3.- Calculate molality

m = moles / mass

m = 1.22/1.12

m = 1.09

b)

Molarity = 0.87      0.87 moles / L

Number of moles = 0.87

mass of solution = density x volume

                           = 1.04 x 1

                            = 1.04 Kg

molality = 0.87 / 1.04

molality = 0.84

c)

Molarity = 5.24 M

Number of moles = 5.24

mass of solution = 1.19 x 1

                           = 1.19

molality = 5.24/1.19

molality = 4.4