13. 91 g of sodium nitrate is dissolved in 100 g of water at 8. 80Cand constant

Question

13. 91 g of sodium nitrate is dissolved in 100 g of water at 8. 80Cand constant pressure.
The enthalpy of dissolving is Hrxn = +?24.5 kJ/mol. How manygrams of ice will be
formed after the mixture reaches thermal equilibrium? The heatcapacity of the solution is
4.18 J/g*oC and the heat of fusion for water isHfus = +?6.006 kJ/mol

(A) 0.2009 g
(B) 0.4509 g
(C) 1. 071 g
(D) 1. 651 g
(E) 0.1509 g

This is what I did (I believe the process is right):
qRXN = -qSOLN
(13.91/80.06)(24500) =-(4.18)(100+13.91)(-8.8-(x/18.02))(-6006)

Explanation / Answer

Yes, what you have done is correct. Just remove the negative mark from before q solution. Heat absorbed by sodium nitrate is equal to heatreleased by water as it freezes to ice. So (13.91 g / 80.06 g/mol) * (24500 J) = [ (4.18 J/g. C) *(113.91 g) * (8.8 C) ]                                                                                   +[( Mass of ice formed / 18.02 g/mol) * 6006 J/mol] solving, we get mass of ice formed = 0.2009 g So the answer is (A)

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