Two beakers are placed in a small closed container at 25 °C. One contains 461 mL
ID: 576353 • Letter: T
Question
Two beakers are placed in a small closed container at 25 °C. One contains 461 mL of a 0.112 M aqueous solution of BaCl2; the second contains 293 mL of a 0.271 M aqueous solution of BaCl2. Small amounts of water evaporate from both solutions. As time passes, the volume of solution in the second beaker gradually................... a) increases or b) decreases
and that in the first gradually.................a) increases or b) decreases
If we wait long enough, what will the final volumes and concentrations be?
461 mL
0.112 M
293 mL
0.271 M
? mL
? M
? mL
? M
First beaker Second beaker Initial461 mL
0.112 M
293 mL
0.271 M
Final? mL
? M
? mL
? M
Explanation / Answer
Ans. If sufficiently long time is given, both the solutions would have the same concertation, i.e. both the solutions have same number of moles per unit volume. When concentration of both the solutions is the same, there is no net movement of water vapor from one to another solution.
Note that only water vapor moves between the beakers but no solute moves from one to another solution.
# Total Volume of both solution = 461.0 mL + 293.0 mL = 754.0 mL = 0.754 L
# Moles of BaCl2 in beaker 1 = Molarity x Volume of solution in liters
= 0.112 M x 0.461 L
= 0.051632 mol
# Moles of BaCl2 in beaker 2 = 0.271 M x 0.293 L = 0.079403 mol
# Total moles of BaCl2 = Moles of BaCl2 in (beaker 1 + Beaker 2)
= 0.051632 mol + 0.079403 mol
= 0.131035 mol
# Overall molarity of if both the solutions are mixed = Total moles / Total volume in L
= 0.131035 mol / 0.754 L
= 0.1738 M
Therefore, the equilibrium concertation after letting the beakers stand for sufficiently long time = 0.1738 M
# Beaker 1 at equilibrium:
We have,
Number of moles of BaCl2 in beaker 1 = 0.051632 mol
Equilibrium [BaCl2] = 0.1738 M
So, required volume of equilibrium solution = Moles / equilibrium molarity
= 0.051632 mol / 0.1738 M
= 0.051632 mol / (0.174 mol/ L)
= 0.2971 L
= 297.1 mL
# # Beaker 2 at equilibrium:
We have,
Number of moles of BaCl2 in beaker 2 = 0.079403 mol
Equilibrium [BaCl2] = 0.1738 M
So, required volume of equilibrium solution = Moles / equilibrium molarity
= 0.079403 mol / 0.1738 M
= 0.079403 mol / (0.174 mol/ L)
= 0.4569 L
= 456.9 mL
# So, As time passes, the volume of solution in the second beaker gradually a. increases (from 293 mL to 456.9 mL) and that is the first gradually b. decreases (from 461 mL to 297.1 mL).
# Equilibrium composition:
Beaker 1: Volume of solution = 297.1 mL ; [BaCl2] = 0.1738 M
Beaker 2: Volume of solution = 456.9 mL ; [BaCl2] = 0.1738 M