Question

The ΔH°soln of HNO3 is –33.3 kJ/mol. 12.0 mL of 11.0 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.

Answer 1

heat released by HNO3 = heat absorbed by water

   DH*no of mol of HNO3 = nwater*swater*DT

33.33*10^3*(12*11/1000) = (100/18)*75.3*(x-25)

   x = Final temperature of solution = 35.51 c

now ice is added to get back to 25 c.

heat absorbed by ice = heat lost by mixture

nice*DHfus+ nwater*swater*DT = nsolution*s*DT

x*6.01*10^3+x*75.3*(35.51-25) = ((12*11/1000)+(100/18))*80.8*(35.51-25)

no of mol of ice must be added = 0.71*18 = 12.78 g

amount of ice must be added= 0.71*18 = 12.78 g