Question

What is the change in enthalpy (in kJ) under standard conditions when 33.75 g of lithium hydroxide dissolves in water? What is the change in enthalpy (in kJ) under standard conditions when 26.58 g of potassium hydroxide dissolves in water? What is the change in enthalpy (in kJ) under standard conditions when 181.37 g of sodium sulfate dissolves in water?

Answer 1

A. Molar mass of lithium hydroxide=Molar mass of Lithium+Molar mass of O+Molar mass of H

=6.94 g/mol+16 g/mol+1 g/mol=23.94 g/mol

Mass of Lithium hydroxide=33.75 g

Number of moles of Lithium hydroxide=mass/molar mass

=33.75 g/23.9 g/mol=1.41 mol

Molar enthalpy of dissolution of LiOH under standard conditions=-23.56 kJ/mol

Enthalpy change of dissolution of given amount of lithium hydroxide=

Molar enthalpy of dissolution x Number of moles

=-23.56 kJ/mol x 1.41 mol=-33.22 kJ

B) Molar mass of potassium hydroxide=Molar mass of potassium+Molar mass of O+Molar mass of H

=39 g/mol+16 g/mol+1 g/mol=56 g/mol

Mass of potassium hydroxide=26.58 g

Number of moles of potassium hydroxide=mass/molar mass

=26.58/56 g/mol= 0.47 mol

Molar enthalpy of dissolution of KOH under standard conditions=- 57.61 kJ/mol

Enthalpy change of dissolution of given amount of lithium hydroxide=

Molar enthalpy of dissolution x Number of moles

=-57.61 kJ/mol x 0.47 mol=-27.08 kJ

C) Molar mass of sodium sulphate=2xMolar mass of sodium+Molar mass of S+4xMolar mass of O

=2x23 g/mol+32 g/mol+4x16 g/mol=46 g/mol+32 g/mol+64 g/mol=142 g/mol

Mass of potassium sulphate=181.37 g

Number of moles of Lithium hydroxide=mass/molar mass

=181.37 g/142 g/mol=1.28 mol

Molar enthalpy of dissolution of Na₂SO₄ under standard conditions=2.4 kJ/mol

Enthalpy change of dissolution of given amount of lithium hydroxide=

Molar enthalpy of dissolution x Number of moles

=2.4 kJ/mol x 1.28 mol=3.07 kJ