Question

PLEASE ANSWER BOTH :)

11. Commercial cold packs consist of solid NH₄NO₃ and water. In a coffee-cup calorimeter, 5.60 g NH₄NO₃ is dissolved in 100.0 g of water at 22.0°C; the temperature falls to 17.9°C. Assuming that the specific heat capacity of the solution is 4.18 J/(g·K), calculate the enthalpy of dissolution of NH₄NO₃, in kJ/mol. Assume that the heat capacity of the calorimeter is negligible.

(1) 24.5 kJ/mol           (2) −24.5 kJ/mol        (3) 25.9 kJ/mol          (4) −25.9 kJ/mol        (5) −1.81 kJ/mol

18. A sample of NaNO₃ was fully decomposed into its elements, and the resulting gas was collected over water at 34°C. If the total pressure of the collected gas was 760 torr, and the vapor pressure of water at 34°C is 40.0 torr, what was the partial pressure of N₂ gas in the flask?

(1) 180 torr                 (2) 190 torr                 (3) 360 torr                 (4) 380 torr                 (5) 720 torr

Answer 1

Part 1)

NH₄NO₃(s)         NH₄⁺(aq) + NO₃^-(aq)

Molar mass of NH₄NO₃(s) = 80.052 g/mol

Moles of NH₄NO₃(s) = Mass/ Molar mass = 5.6 g/(80.052 g/mol) = 6.99 x 10^-2 moles

The total mass of NH₄NO₃ and water = (5.6 + 100) g = 105.6 g.

The heat change is given by,

q = c x m x T

Heat capacity of water = 4.18 JK^-1g^-1

Therefore,

q = 4.18 x 105.6 x (22-17.9) = 1809 J = 1.809 KJ

This is the heat change produced by 6.99 x 10^-2 moles

Therefore,

Enthalpy of dissolution of NH₄NO₃ = (1.809 KJ)/( 6.99 x 10^-2 moles) = +25.9 KJ/mol

Part 2)

2NaNO₃   2Na + N₂ + 3O₂

Therefore, if P is the pressure of N₂ produced, pressure of O₂ produced will be 3P

Here, P_total = 760 torr, also P_{water vapour} = 40 torr

Therefore,

P_N2 + 3P_O2 + P_H2O = 760 torr

4P = (760-40)torr = 720 torr

P = 180 torr

Partial pressure of N₂ gas in the flask = 180 torr