8.43g of cacl2·2h2o is dissolved in 40.0g of h2o. what is the molality of the solution? (note that the h2o in the cacl2·2h2o must be included in the molar mass of the compound in order to calculate the moles of solute. also note that the h2o contributed by this solute must be included in the total amount of solvent present in the solution in calculation of molality.

[CaCl₂·2H₂O] = 1.43 m

Explanation:

Molality is mol of solute / kg of solvent.

Mass of solvent = 40 g

Let's convert g to kg → 40 g / 1000 = 0.04 kg

Let's determine the moles of solute (mass / molar mass)

8.43 g / 146.98 g/mol = 0.057 mol

Molality = 0.057 mol / 0.04 kg → 1.43

The answer to your question is molality = 1.36 ≈ 1.4

Explanation:

Data

mass CaCl₂ 2H₂O= 8.43 g

mass of water = 40 g

molality = ?

Process

1.- Calculate the molecular mass of CaCl₂ 2H₂O

CaCl₂ 2H₂O = 40 + 71 + 4 + 32 = 147 g

   Calculate the moles of CaCl₂ 2H₂O

                          147 g 1 mol

                            8.43 g  x

                             x = (8.43 x 1) / 147

                             x = 0.057 moles

2.- Calculate the mass of H₂O in the solute

                        147 g of CaCl₂ 2H₂O  36 g of H₂O

                          8.43 g CaCl₂ 2H₂O   x

                          x = (8.43 x 36) / 147

                          x = 303.48 / 147

                          x = 2.06 g of H₂O

3.- Total mass of water

     mass of water = 2.06 + 40

                              = 42.06 g

4.- Calculate molality

     molality = number of moles / mass of solvent (kg)

     molality = 0.057 / 0.0421

     molality = 1.4

the only liquid elements at standard temperature and pressure are bromine (br) and mercury (hg). although, elements caesium (cs), rubidium (rb), francium (fr) and gallium (ga) become liquid at or just above room temperature.

explanation: