Question

7) (10 pts) In the titration of sodium oxalate with potassium permanganate, very pure sodjum oxalate (Na2C2O4, MM133.9985 g/mol) is used to find an accurate Molarity of a potassium permanganate (KMnO) solution. If 98.52 mg of sodium oxalate is dissolved in exactly 25.00 mL of solution at 25.0°C, 21.50 mL of KMnO solution is needed to achieve equivalence. What is the Molarity of the KMnO,? Sodium and potassium ions are spectators in this reaction. Constant normal Ts and Ps are observed. 5 C20.-(aq) + 2 MnO, (aq) + 16 H(aq) → 10 CO, (aq) + 2 Mn? (aq) + 8 H,OM 8) (6 pts) What is the kinetic energy in J) of an oxygen molecule moving at a speed of 589 m/s? K.E. = Vamy 6.022 x 10" "entites"/mol 9) (6 pts) At 0.998 atm pressure, an originally empty hot air balloon is filled to 2.50 x 10° L, and rises to 25,000 ft., absorbing 365,000 kJ of heat. What is the Total Energy (kJ) for the process? 101.325 J = 1.00 L-atm AE = q + w w--PAV

Answer 1

7) Fro the stoichiometry of the reaction we can see that per each 2 moles of MnO₄^- added, 5 moles of oxalate react. The original number of moles of oxalate present in the solution is given by:

$n=\frac{m}{m_{molar}}=\frac{0.09852g}{134g/mol}=7.352x10^{-4}moles$

And the number of moles of permanganate necessary to react with these is:

$n_{MnO_{4}^{-}}=\frac{2}{5}\cdot 7.352x10^{-4}moles=2.941x10^{-4}moles$

These moles are present un a volume of 21.50 mL, so the concentration of potassium permanganate is

$M=\frac{n}{V(L)}=\frac{2.941x10^{-4}moles}{0.02150L}=0.01368M$

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8) To calculate the kinetic energy we need the velocity of the molecule (we already know it) and it's mass. We know that a mole of oxygen molecules weights 32 g and that there are 6.022x10²³ molecules is a mole, so the mass of a single molecule is;

$m=\frac{0.032kg/mol}{6.022x10^{23}molecules/mol}=5.314x10^{-26}kg/molecule$

We can now calculate the kinetic energy using the given formula:

$KE=\frac{1}{2}\cdot m\cdot v^{2}=\frac{1}{2}\cdot 5.314x10^{-26}kg\cdot (589\frac{m}{s})^{2}=9.22x10^{-21}J$

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9) The change in internal energy, as expressed in the exercise, is given by the addition of heat and work. In this case, we can calculate them pretty directly (we will assume that the initial volume of the empty balloon is zero)

$\Delta E=q-p\Delta V=365,000,000J-0.998atm\cdot (2.50x10^{6}L-0)\cdot \frac{101.325J}{1L\cdot atm}=112194125J=112,124kJ$