Question

Americans spend up to $100 billion annually for bottled water (41 billion gallons). The only beverages with higher sales are carbonated soft drinks. Recent news stories have highlighted the fact that most bottled water comes from municipal water supplies (the same source as your tap water), although it may undergo an extra purification step called reverse osmosis. Imagine two tanks that are separated by a membrane that is permeable to water, but not to the dissolved minerals present in the water. Tank A contains tap water and tank B contains the purified water. In the reverse osmosis process, pressure is applied to the tap water tank to force the water molecules across the membrane into the pure water tank. Suppose that while the reverse osmosis system is running tank B becomes contaminated with salt water such that it ends up with a higher salt concentration than tank A. What would happen if the reverse osmosis system was turned off?

Answer 1

Answer: C) be hypertonic to Tank B.

Explanation:

The ability of an extracellular solution to move water in or out of a cell by osmosis is known as its tonicity. Additionally, the tonicity of a solution is related to its osmolarity, which is the total concentration of all the solutes in the solution.

Three terms (hypothonic, isotonic and hypertonic) are used to compare the osmolarity of a solution with respect to the osmolarity of the liquid that is found after the membrane. When we use these terms, we only take into account solutes that can not cross the membrane, which in this case are minerals.

If the liquid in tank A has a lower osmolarity (lower concentration of solute) than the liquid in tank B, the liquid in tank A would be hypotonic with respect to the latter.

If the liquid in tank A has a greater osmolarity (higher concentration of solute) than the liquid in tank B, the liquid in tank A would be hypertonic with respect to the latter.

If the liquid in tank A has the same osmolarity (equal concentration of solute) as the liquid in tank B, the liquid in tank A would be isotonic with respect to the latter.

In the case of the problem, option A is impossible because the minerals can not cross the membrane, since it is permeable to water only. There is no way that the concentration of minerals decreases in tank A, so the solution in this tank can not be hypotonic with respect to the one in Tank B.

Equally, both solutions can not be isotonic and neither we can say that the solution in tank A has more minerals that the one in tank B because the liquid present in tank B is purified water that should not have minerals. Therefore, options B and D are also not correct.

Finally, the correct option is C, since in the purification procedure the water is extracted from the solution in tank A to obtain a greater quantity of purified water in tank B. In this way, the solution in Tank A would be hypertonic to Tank B.

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