Answer to Question #150907 in General Chemistry for Lyna Moinifar

Question

When 50.00 mL of aqueous HCl was mixed with 50.00 mL of NaOH (in large excess), the temperature of the solution increased from 25.00 °C to 32.40 °C. The reaction is
NaOH(aq) + HCl(aq) ↔ NaCl(aq) + H 2O(aq) -- ΔH = -57.3 kJ
What was the molarity of the original HCl solution? Assume the heat capacity of the solution is the same as pure water (4.184 J/g*°C), the density of the solution is 1.00 g/mL and there is no loss of heat to the surroundings.

Key Concept:: Calorimetry and enthalpy change is an extensive property. V, d → m solution; ΔT, heat capacity → heat; ΔH→ mol of HCl; V → [HCl]

Answer 1

∆H = nC∆T

-57300 = n * 4.18 *36.5 * (32.40-25)

-57300 = n * 4.18*36.5*7.40

57300 = n× 1129.01

n = 57300/1129.01

n = 50 moles

molarity = no. of moles/volume

molarity = 50moles/0.05L = 1000M