What is the enthalpy change(Q) when 60.0 g of NaOH is dissolved in one litre of water, given that the temperature of the solution increased by 15.8 °C?

ΔH = -44.029 KJ/mol

Explanation:

We are told that 60.0 g of NaOH is dissolved in one litre of water.

From conversion, 1 litre = 1000 g

From tables, specific heat capacity of water; c = 4.18 J/g/°C

We are given ΔT = 15.8 °C

Formula for the heat absorbed by the solution is given as;

Q = mcΔT

Where;

Q = Heat gained by the water

m = Mass of the water

c = Specific heat of water

ΔT = change in temperature

Thus;

Q = 1000 × 4.18 × 15.8

Q = 66044 J

The solution absorbed 66044 J and therefore it means that the dissolution of the salt gave off 66044 J.

Thus;

Enthalpy is; ΔH = -66044 J

Now, Mass of NaOH is 60 g.

We know that molar Mass of NaOH is 39.997 g/mol. Thus;

Converting to moles, we have;

n = 60/39.997

n = 1.5 mol

Now, enthalpy when 1.5 moles of NaOH are dissolved in water will be;

ΔH = -66044 J/1.5 moles

ΔH = -44029.33 J/mol

ΔH = -44.029 KJ/mol

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.