Question

10.23 g of sodium thiosulfate pentahydrate (ΔHsoln=+47.40 kJ/mol; MM = 158.108 g/mol) is dissolved in water. The mass of the water solution is 102.5 g. If the initial temperature of the water was 22.8°C, what will be the final temperature of the solution?

Answer 1

1 mol No of moles of sodium thiosulfate pentahydrate = 10.23g xe - = 0.0647 mol 158.108 g Heat of solution = –15.30 kJ/mol +47.40 kJ Amount of heat absorbed in the reaction = 0.0647 mol x 1 mol = +3.067 kJ = + 3067 J Amount of heat asorbed in the reaction is equal to the heat released by water We known that, Iwater =-4mm =-3067 J Mass of water, m =102.5 g Specificheat of water ,s = 4.184 J/g.°C Initi al temperature 77= 22.8°C Amount of heat absorbed by water, qarater = ms(7-7) T3-T; = I were -3067 J 102.5 g *4.184 – - = -7.15°C gºC Final temperature, T = -7.15ºC + T = -7.15ºC+22.8°C = 15.65°C = 15.6°C