Q = 6.00 kJ
given that the heat capacity of the final solution is the same as that of pure water of equal mass.
ΔH = 468 kJ/mol.
Both answers come with three sig. fig. as in data in the question.
Final mass of the solution:
[tex]m = \text{Mass of Solution} = \text{Mass of Solute} + \text{Mass of Solvent} = 51.55 + 5.12 = 56.67\;\text{g}[/tex].
An increase in temperature by one degree celsius is the same as an increase in temperature by one degree Kelvin. Change in temperature:
[tex]\Delta T = \text{Final Temperature} - \text{Initial Temperature} = 49.8 - 24.5 = 25.3\;\text{K}[/tex].
Assume that the specific heat capacity of the NaOH solution is the same as the specific heat capacity of pure water:
[tex]c = 4.182\;\text{J}\cdot\text{g}^{-1}\cdot\text{K}^{-1}[/tex].
In other words, it takes 4.182 Joules of energy to raise the temperature of one gram of pure water by one degree Kelvin. Assume that so is the case for the NaOH solution.
Energy change:
[tex]Q = c\cdot m \cdot \Delta T = 4.182 \times 56.67 \times 25.3 = 5995.95\times 10^{3} \;\text{J} = 6.00\times 10^{3}\;\text{J}[/tex].
Refer to a periodic table for relative atomic mass:
Formula mass of NaOH:
[tex]M(\text{NaOH}) = 22.990 + 15.999 + 1.008 = 39.997\;\text{g}\cdot\text{mol}^{-1}[/tex].
How many moles of formula units in that 5.12 gram of NaOH?
[tex]\displaystyle n = \frac{m}{M} = \frac{5.12}{39.997} = 0.128010\;\text{mol}[/tex].
ΔH measures the energy change per mole of NaOH dissolved. Dissolving 5.12 grams or 0.128010 moles of NaOH releases [tex]5.99595\times 10^{3} \;\text{J}[/tex] of energy.
[tex]\displaystyle \Delta H = \frac{Q}{n} = \frac{5.99595\times 10^{3}\;\text{J}}{0.128010\;\text{mol}} = 4.68\times 10^{5}\;\text{J}\cdot\text{mol}^{-1} = 4.68\times 10^{2}\;\text{kJ}\cdot\text{mol}^{-1}[/tex].
