Question

You add 120.0 mL of water at 4.0 C to 2.50 moles of

anhydrous barium chloride solid. The hydrate that anhydrous barium
chloride forms is barium chloride tetrahydrate.
a. Write the balanced reaction equation for hydration reaction. You
may assume that only unit cells of barium chloride tetrahydate form in
the reaction.
b. Calculate the moles of hydrate formed when the water is added.
c. What percentage of the sample is not fully hydrated?
d. You add an additional 650.0 mL of water at 4C and all of the
solute dissolves. Calculate the density of the resulting solution, if the
total solution volume is 795.0 mL after all of the water has been added.

Answer 1

a. The balanced reaction equation for the hydration of anhydrous barium chloride to form barium chloride tetrahydrate is:

BaCl2 (s) + 4 H2O (l) → BaCl2.4H2O (s)

b. The amount of hydrate formed when the water is added can be calculated by using the stoichiometry of the reaction equation. Since 4 moles of water are used for every 1 mole of anhydrous barium chloride, we can calculate the moles of hydrate formed as follows:

moles of hydrate = (moles of water) / (moles of anhydrous BaCl2) * (moles of hydrate)

moles of hydrate = (120.0 mL * 1L/1000mL * 1 mole/18.015 g * 1 g/mL) / (2.50 moles) * (1 mole hydrate / 1 mole anhydrous BaCl2) = 0.03 moles

c. To find the percentage of the sample that is not fully hydrated, we can use the following formula:

(moles of anhydrous BaCl2 - moles of hydrate) / (moles of anhydrous BaCl2) * 100% = (2.50 moles - 0.03 moles) / 2.50 moles * 100% = 98.8%

d. To find the density of the resulting solution, we can use the following formula:

density = mass/volume

mass = moles of solute * molar mass of solute + moles of solvent * molar mass of solvent

mass = (2.50 moles * 208.23 g/mole) + (650.0 mL * 1L/1000mL * 1g/mL) = 518.1g + 650g = 1168.1g

volume = 795.0 mL * 1L/1000mL = 0.795L

density = mass/volume = 1168.1g / 0.795L = 1473g/L

The density of the resulting solution is 1473g/L