The ΔH°soln of HNO3 is –33.3 kJ/mol. 17.0 mL of 13.0 M HNO3 is dissolved in...

The ΔH°soln of HNO3 is –33.3 kJ/mol. 17.0 mL of 13.0 M HNO3 is dissolved in 100.0 mL of distilled water initially at 25°C. How much ice at 0°C [Cp = 37.1 J/(mol ·°C), ΔH°fus = 6.01 kJ/mol] must be added to return the solution temperature to 25°C after dissolution of the acid and equilibrium with the ice is reached? The molar heat capacity is 80.8 J/(mol·°C) for the solution, and the molar heat capacity is 75.3 J/(mol·°C) for pure water.

Expert Solution

no of mol of HNO3 dissolved = 17*13/1000 = 0.221 mol

heat liberated(q) = -33.3*0.221 = -7.36 kj

heat liberated(q) = heat gained by water + HNO3 solution

= n*Cm*DT

7.36*10^3 = (100/18)*75.3*(x-25) + 0.221*80.8+(x-25)

final temperature of solution = x = 42.51 C

heat gained by ice = heat lost by solution

(x/18)*37.1*(25-0)+(x/18)*6.01*10^3 = (100/18)*75.3*(42.51-25) + 0.221*80.8+(42.51-25)

x = amount of ice must be added = 19.1 g

queen_honey_blossom answered 2 years ago

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