An aqueous solution is made by dissolving 26.4 grams of barium sulfide in 349 grams of water. The molality of barium sulfide in the solution is
m.

Molality = 0.43 m

Explanation:

Given data:

Mass of barium sulfide = 25.4 g

Mass of water = 349 g (349 g/1000 = 0.349 Kg)

Molality of water = ?

Solution:

Number of moles of barium sulfide:

Number of moles = mass/molar mass

Number of moles = 25.4 g/169.39 g/mol

Number of moles = 0.15 mol

Molality:

Molality = number of moles o f solute/ Kg of solvent

Molality = 0.15 mol / 0.349 Kg

Molality = 0.43 m

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%

explanation:

0.07 g/cm³

1.37 g/cm³