In a second experiment, 16.00 g of sodium acetate trihydrate (136.08 g/mol) was dissolved in 100.00...

In a second experiment, 16.00 g of sodium acetate trihydrate (136.08 g/mol) was dissolved in 100.00 ml of water. The initial temperature was 25.00 °C, and the final temperature was 20.06 ºC. The heat capacity of the reaction mixture was 4.045 J g^-1deg^-1

E. Calculate delta H_hydration for converting sodium acetate(s) to sodium acetate trihydrate (aq)

I've done previous portions of this same question including calcultating delta T, heat transfered, number of moles, and delta H _dissolution if that helps to know!

Solutions

Expert Solution

mass of solution = 16.00+ 100 = 116 g

q = msdeltaT

= 116*4.045*(20.06-25)

= - 2317.95 J

moles of sodium acetate trihydrate = 16.00/136.08 = 0.1176 moles

0.1176 moles of sodium acetate gives ----> -2317.95 J

1 mole of sodium acetate gives ------> -2317.95*1/0.1176

= -19710.46 J

= -19.71046 KJ/mol

queen_honey_blossom answered 1 year ago

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