Question

You have four beakers (A, B, C, and D)

with different concentrations of sugar water in
them. The
concentrations in the beakers are 10% sugar,
15% sugar, 20% sugar, and 30% sugar, but
these
concentrations are not necessarily in the
order A, B, C, and D. (In other words, beaker
A does not
necessarily have the 10% sugar solution in
it.) You also have 4 pieces of dialysis tubing,
each one with a
mass of 10 grams of sugar water.
The sugar water in the dialysis tubes is 20% sugar. You
place a piece of
tubing in each of the beakers and leave it
for 2 hours. At the end of 2 hours you remove the
tubes and
mass them again. Assume the dialysis tubing
is permeable to water but is not permeable to
sugar. The
table below summarizes the data.
Beaker
Initial Mass of Dialysis Tube
Mass of tube after 2 hours
A
10.0g
11.4g
B
10.Og
10.0g
с
10.0g
13.2g
D
10.0g
8.8g
Based on this information, determine which beaker contained the 10% sugar solution,
which beaker
contained the 15% sugar solution, which beaker contained the 20% sugar solution, and which beaker

contained the 30% sugar solution (4 points)
"

Answer 1

Beaker A contained the 10% sugar solution, as water moved into the comparatively hypertonic 20% sugar solution in the dialysis tubing. Beaker B contained the 20% solution due to no mass change, suggesting an isotonic state. Beakers C and D had to be the 15% and 30% solutions, respectively, with C having lower concentration than the tubing's and D having a higher concentration.

Step-by-step explanation:

According to the principles of osmosis, water moves across a semipermeable membrane from an area of lower solute concentration (hypotonic) to an area of higher solute concentration (hypertonic). The dialysis tubing in the experiment is only permeable to water, not sugar, resulting in changes in the mass of the tubing due to the movement of water in or out of the tubing.

The tubing in beaker A gained mass, indicating that water moved into the tubing, which means beaker A contained a solution that was more dilute than the 20% sugar solution in the tubing. Therefore, beaker A contains the 10% sugar solution. The tubing in beaker B remained the same, suggesting that the concentration inside the tubing and the beaker was in equilibrium; hence, beaker B contained the 20% sugar solution.

Beaker C's tubing gained even more mass than A's, indicating that beaker C was even more dilute than A and was pure water. Therefore, beaker C had no sugar, but as it is not one of the options, we disregard this in favor of the known concentrations. The tubing in beaker D lost mass, signifying that water moved out of the tubing into a hypertonic solution, which means that beaker D contained the 30% sugar solution. Following the process of elimination, beaker C, which had the highest water mass gain, contained the 15% sugar solution.