Practice Passage Test - 3 | Practice Passages for MCAT PDF Download

The passage states that volatile chemicals cannot be used in samples subjected to lyophilization because they will hinder sublimation and favor the liquid phase through melting. This is because these compounds have a high vapor pressure and cannot be removed from the system by the condenser. Choices A, B, and D are all hydrogen donors and acceptors. Choice C, diethyl ether, can only act as a hydrogen bond acceptor. Therefore, it has the weakest intermolecular forces and is the most volatile.

On a P vs. T diagram, sublimation equilibrium is indicated by the line separating the solid and vapor phases. On the graph in Figure 1, it is shown that at -60°C, the solid/gas line is at P = 0.0003 or 3 × 10^-4 atm.

Both dichloromethane and methanol are removed as liquids at reasonably low temperatures and have freezing points far lower than water. This eliminates choices A and D. The passage states that one of the major advantages to lyophilization is that it can be performed at low temperatures, and therefore preserve the activity of heat-sensitive samples. NaCl is a salt that is stable at high temperatures, whereas enzymes are proteins that denature at high temperatures. Therefore, lyophilization is best suited as a means to remove water from aqueous protein solutions.

Sublimation is an endothermic reaction, requiring heat input. As this reaction removes heat from the surroundings, it lowers the temperature of the surroundings, in this case, the reaction chamber. This is very similar to sweat cooling the body as it evaporates off of the skin (also an endothermic process).

40 cm³ of ice is 36 g or 2 moles of water. The heat required to sublimate this sample is 46 kJ/mol(2 mol) = 92 kJ. If heat is transferred at 0.1 kJ/min, then 920 minutes are required. Dividing 920 min by 60 min/hour gives just over 15 hours.