A 2.50 g sample of solid sodium hydroxide is added to 55.0 mL of 25 °C water in a foam cup (insulated from the environment) and stirred until it dissolves. What is the final temperature of the solution? ΔHsoln = -44.51 kJ/mol.

37.1°C.

Explanation:

Firstly, we need to calculate the amount of heat (Q) released through this reaction:

∵ ΔHsoln = Q/n

no. of moles (n) of NaOH = mass/molar mass = (2.5 g) / (40 g/mol) = 0.0625 mol.

The negative sign of ΔHsoln indicates that the reaction is exothermic.

∴ Q = (n) (ΔHsoln) = (0.0625 mol) (44.51 kJ/mol) = 2.78 kJ.

We can use the relation:

Q = m. c.ΔT,

where, Q is the amount of heat released to water (Q = 2781.87 J).

m is the mass of water (m = 55.0 g, suppose density of water = 1.0 g/mL).

c is the specific heat capacity of water (c = 4.18 J/g.°C).

ΔT is the difference in T (ΔT = final temperature - initial temperature = final temperature - 25°C).

∴ (2781.87 J) = (55.0 g) (4.18 J/g.°C) (final temperature - 25°C)

∴ (final temperature - 25°C) = (2781.87 J) / (55.0 g) (4.18 J/g.°C) = 12.1.

∴ final temperature = 25°C + 12.1 = 37.1°C.