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Topic: Hess' Law and Calorimetry  (Read 2580 times)

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Offline mgong41

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Hess' Law and Calorimetry
« on: November 07, 2011, 08:04:42 PM »
4.00 g of NaOH (s) is dissolved in 180 mL of pure water (density =0.99 g/mL) at an initial temperature of 19.5 degrees Celcius.  The final temperature is 25 degrees Celcius and the density of the final solution was 1.013 g/mL.  The solution specific heat wsa 4.08 J/g-deg.  The calorimeter constant is 21 J/deg.

1) What is the molar concentration of the final solution?

Is this just referring to molarity? (4.00 g/ 180 mL)(1000 mL/ 1 L) = (22.2 g/L) which is the molarity

(in the hint it says to remember that a solution is made up of a solute and solvent so the mass of the solution is necessary. Why?)

2) What is the molar enthalpy of the dissolution?

Does this involve the q=mC(change in T) equation? If so, would it be:
q=(184.0 g)(4.08 J/g-deg)(5.5 deg)

q=4128.96 J ?

Why are so many unnecessary values given?

Thank you

Offline Vidya

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Re: Hess' Law and Calorimetry
« Reply #1 on: November 07, 2011, 10:09:33 PM »
a)molarity is the moles of the solute per liter of the solution
hence mass of the solution is must
you convert total mass of the soln into volume from the density of the solution.Now you can use this volume in Liter to get the molarity of the solution 

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