Question

Based upon the coffee cup calorimeter calculations, solve the following:

0.962g of NH4Cl is dissolved in 20 ml of H2O in a coffee cup calorimeter. The temperature of the solution decreases from 24 to 22o C. Calculate the enthalpy change for the dissolution

of NH4Cl in KJ/mole (Hint: Molar mass of NH4Cl is 53.491g/mole) .

Answer 1

Puye solution - Tissotution of N Hych: NHC - बोन ) AH = ? mess of NH y cl = 0.962 g molar mass of N Hy cl = 53.491 g/mol, mol of NH i ce a given mess molan mess 0.962 g 53.491 g/mol = 0.01798 mol. now, volume of water = 20 ml density of water = 1 gn/me mass of water = volume x dersity => mw=20mex I g/me - ² mw = 2ogm. initial temperature of water, T = 24°C tinal temperature of Water, Tz = 22°C. pecetic head of water, Cp (H2O (1)) = 4.18 J/g.°c here, temperature of Water decreased. so, heat is released by the water. - qw = mw x Cp (H2O(A)) X AT = mw X cp (H2(e)) x (Tz -T) = 20 gm x 4:18 / 4X (22-244 qw = - 167.2 57 mmmmm W OXES

e Page - Page minus sign indicate heat is released by the water. This amount of heat will be absorbed by NHyl. now, heat absorbed by N Hy l = mol A N H y a X 4H. => INH.ce = 0.01798 X 4H . mo ow heat absorbed by Nhoc - heat de les by water =) Nhy Cl = -1 = 0.01798 x 4H = -(-16712) J. => AH = 16+ 0.01798 mol mol = 9299.22 I/mol = 9:299 kJ/mo), t. N Hy l = Ntyt tar (44) ray) AH = 9. 299 KI/mol. change for the dissolution of 9' 299 kJ/mol enthalpy Nhy I is nor