[SOLVED]

When 15.3g of sodium nitrate, NaNO3, was dissolved in water in a constant-pressure calorimeter, the temperature fell from 25.00 degrees Celsius to 21.56 degrees Celsius. If the heat capacity of the solution and the calorimeter is 1071 Joules per degrees Celsius, what is the enthalpy change when 1 mol of sodium nitrate dissolves in water?

The solution process is

NaNO3(s) ---> Na+(aq) + NO3-(aq) ; Enthalpy change = ?

---------------------------------------...

I decided to use the equation q = (mass) x (specific heat) x (change in temperature)

I divided 1071 Joules per degrees Celsius by 15.3g of NaNO3 to get 70 Joules per grams x degrees Celsius and I used that as the specific heat for the equation:

q = (15.3 g) (70 J/g*C) (-3.44 C)

q = -3684.24 J

I know I am doing something wrong... but I am confused now. I need a chemistry expert help!

The answer at the back of the book is : 20.5 kJ

When 15.3g of sodium nitrate, NaNO3, was dissolved in water in a constant-pressure calorimeter, the temperature fell from 25.00 degrees Celsius to 21.56 degrees Celsius. If the heat capacity of the solution and the calorimeter is 1071 Joules per degrees Celsius, what is the enthalpy change when 1 mol of sodium nitrate dissolves in water?

The solution process is

NaNO3(s) ---> Na+(aq) + NO3-(aq) ; Enthalpy change = ?

---------------------------------------...

I decided to use the equation q = (mass) x (specific heat) x (change in temperature)

I divided 1071 Joules per degrees Celsius by 15.3g of NaNO3 to get 70 Joules per grams x degrees Celsius and I used that as the specific heat for the equation:

q = (15.3 g) (70 J/g*C) (-3.44 C)

q = -3684.24 J

I know I am doing something wrong... but I am confused now. I need a chemistry expert help!

The answer at the back of the book is : 20.5 kJ

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