An Exercise in Freezing Point Depression

Sample Calculation: Determine the Enthalpy of Solution of Ammonium Chloride.A student is asked to add ~2g of NH 4Cl to 10mL of water and to use the change intemperature to calculate ∆H soln. The reaction below corresponds to the enthalpy ofsolution of ammonium chloride.NH 4 Cl(s) + H 2 O(l) → NH 4 + (aq) + Cl - (aq)∆H solnA student first checked the solubility of NH 4Cl to make sure that a 2 g sample woulddissolve in 10 mL of water. She reasoned that because 2.97 g will dissolve in 10 mL ofwater at 0°C and since the solubility of NH 4Cl increases with temperature, a 2 g sampleshould readily dissolve at room temperature. She then stirred 2.105 g of NH 4Cl into 10.00mL of DI water and observed that the temperature changed from 24.2°C to 11.1°C. Thecalorimetry constant for the well used was 5.8 J/°C. If q solnis the heat transfer associatedwith the dissolving of the salt sample and q is the heat lost or gained by the solution ofwater and salt, thenq soln + q + q cal = 0q soln + (mass of salt + mass of water)(C sp solution)(∆T) + (5.8 J/°C)(∆T) = 0q soln + (2.105g NH 4 Cl + 10.0 mLx 0.997g/mL)(3.31 J/g°C)(11.1°C – 24.2°C) +(5.8 J/°C)(11.1°C – 24.2°C) = 0q soln − 524 J − 76 J = 0 and q soln ≅ 6.00 x 10 2 J∆H(J/g) = (6.00 X10 2 J)÷(2.105 g) = 285 J/g∆H (kJ/mol) = (285 J/g) (1 kJ/ 1000 J) (53.49g/mol) = 15.2 kJ/mol10-12