A solution is made by dissolving 37.5 g of sodium sulfide (Na2S) in

217 g of water.

What is the molality of the solution?

2.21m

Explanation:

Molality, m, is defined as the ratio between moles of solute (The substance that is in less amount, sodium sulfide) and kilograms of solvent (Water, in this case).

Thus, we need to convert mass of sodium sulfide to moles using its molar mass and grams of water to kilograms:

Moles Na2S (Molar mass: 78.0452g/mol):

37.5g * (1mol / 78.0452g) = 0.48 moles Na2S

Kilograms water:

217g * (1kg / 1000g) = 0.217kg

Molality:

0.48 moles Na2S / 0.217kg =

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%