Question

SUUXOM La(10+5), Soluhall can be prepare (10 pts) What is the Molarity (after mixing) of the CT ions in solution when 25.00 mL of an aqueous 1.055 M KCl solution is mixed with 50.00 mL of an aqueous 0.6245 M BaCl2 solution? T and P are normal and constant; volumes are additive.

Answer 1

6) The best way to do this is to first calculate the total number of moles of Cl^- in the mixture and then take into account the volume to calculate the concentration.

In the KCl solution the number of moles of Cl^- is the same as the number of moles of KCl, since there is one Cl^- per each KCl. The number of moles is:

$n=M\cdot V(L)=1.055\frac{mol}{L}\cdot 0.02500L=0.02637moles$

For the BaCl₂ solution, the concentration of Cl^- is double the concentration of salt, since there are two Cl^- per each BaCl₂. The number of moles is thus:

$n=2\cdot M\cdot V(L)=2\cdot 0.6245\frac{mol}{L}\cdot 0.05000L=0.06245moles$
The total number of moles is:

$n_{total}=0.02637moles+0.06245moles=0.08882moles$

These moles are dissolved in a final volume of 75.00 mL (25.00 mL + 50.00 mL), so their total concentration is given by:

$M=\frac{n}{V(L)}=\frac{0.08882moles}{0.07500L}=1.184M$

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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)

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