Question

An aqueous solution is made by dissolving 25.0 grams of sodium sulfide in 317 grams of water. The molality of sodium sulfide in the solution is m?

Answer 1

1.01 m

Explanation:

Molality is defined as the number of moles of solute divided by the mass of the solvent (in kg). To find the number of moles, use the molar mass of sodium sulfide (`Na_2 S`) by adding up the mass numbers from the periodic table:

`23 + 23 + 32 = 78` g/mol

Then, divide the given mass of `Na_2 S` by its molar mass to get the number of moles:

`25/78 = .32` mol

Finally, divide the number of moles of `Na_2 S` by the mass of the solvent (in this case water):

`.32/.317 = 1.01` m