Question

A sample of Na2SO4(s) is dissolved in 233 g of water at 298 K such that the solution is 0.298 molar in Na2SO4. A temperature rise of 0.129°C is observed. The calorimeter constant is 330. J⋅K−1. Calculate the enthalpy of solution of Na2S04 in water at this concentration.The density of the solution can be approximated to be the density of pure water at 298 K (0.997 g/mL).CP,m(H2O)=75.3J⋅mol^−1⋅K^−1.

Answer 1

The enthalpy of solution of Na2S04 in water is -2.415 * 10³ J/mol

Step-by-step explanation:

Step 1: Given data

The water has a mass of 233 grams and a density of 0.997 g/mL

⇒volume of water is 233 grams / 0.997 g/ mL = 233.7 mL = 0.2337 L

This makes this a 0.298 molar solution

The temperature rise = 0.129°

The calorimeter constant = 330 J/K

Cp,m(H2O) = 75.3 J/mol*K

Step 2: Calculate the enthalpy of solution

m/M *ΔHsolution,m + mH20/MH20 * CH20,m * ΔT + Ccalorimeter ΔT = 0

This we can rearrange to:

ΔHsolution, m =- ((233/18.02)*75.3 * 0.129 + 330 * 0.129) /(0.2337*0.298)

ΔHsolution, m = -2.415 * 10³ J/mol

The enthalpy of solution of Na2S04 in water is -2.415 * 10³ J/mol