The heat of solution of NaOH is -44.5 kJ/mol NaOH. find maximum temperature, originally at 21°C, be raised in the preparation of 500 mL of 7.0 M NaOH? Has density of 1.08 g/mL and heat of 4.00 J/g°C.

Answer 1

The way your question is phrased makes little sense, so I've taken the liberty of rearranging the information, and guessing at what the question is. Here's my take...

∆H = -44.5 kJ/mol

Initial temperature = 21ºC

Final temperature = ?

Specific heat = 4.00 J/gº

mass of solution = 500 ml x 1.08 g/ml = 540 g

moles NaOH = 0.5 L x 7.00 mol/L = 3.5 moles

q = heat = 3.5 moles x -44.5 kJ/mol = -155.8 kJ

q = mC∆T where q = heat; m = mass; C = specific heat; ∆T = change in temperature

155.8 kJ = (540 g)(0.00400 kJ/gº)(∆T)

∆T = 72.1º

Since q is negative, this means the reaction is exothermic and thus the temperature of the solution will increase. Thus, the final temperature will be 21º + 72.1º = 93ºC