Question

6 Question (1point) See page 228 Adding nitric acid to water is quite exothermic. Calculate the temperature change (AT) when 100.0 mL of water (d=1.00 g/mL) at 20.0°C [c-75.3 J/(mol-°C)] and 10.0 mL of concentrated HNO(16.0 M,AHin-33.3 kJ/mol) are combined. 1st attempt See Periodic Table See Hint Assume the concentrated HNO3 has a density equal to that of water and its addition doesn't add a significant amount of water to the system. °C

Answer 1

Ans :

Mol HNO3 = molarity x volume (L)

= 16.0 M x 0.010 L

= 0.16 mol

heat released = mol x Hsol

= 0.16 mol x -33.3 KJ/mol

= -5.328 KJ

= -5328 J

this heat will be gained by the solution

mol water = mass / molar mass

= 100 g / 18.01528 g/mol

= 5.55 mol

heat = mole x specific heat x change in temperature

5328 J = 5.55 mol x 75.3 J/(mol.^oC) x T

T = 12.7^oC

so temperature change is 12.7^oC