Use the spaces provided for the answers and additional paper if necessary) 1. A

Question

Use the spaces provided for the answers and additional paper if necessary) 1. A student was given a sample of an anhydrous salt to analyze, using the procedure described in this experiment. The student weighed the sample but decided to do the experiment the following day. The weighed sample was left uncovered on the bench overnight. (1) What error(s) might result if the student used this salt sample in a determination the following day? Briefly explain. (2) Would the student avoid any error(s) described in (1) by reweighing the salt sample on the following day, before doing the experiment? Briefly explain what effect the use of a glass beaker in place of the pressed polystyrene cup would have on the expected (3) hydration.

Explanation / Answer

(1). As the sample of uranhydrous salt was left over uncovered in the lab for overnight, If the salt sample was originally very dry and also hygroscopic, it would probably have reacted with some of the moisture in the air and this would change the measured enthalpy of hydration for the experiment. As error will add on to the weight if compared with the original state.

Some anhydrous salts can absorb moisture from the air to become hydrated and called as hygroscopic salts.Therefore some hygroscopic salts like silica gel and are useful as drying agents. Which is usually packaged with optical or electronic devices shipped by boat. The silica gel protects the devices from high humidity. There are some hygroscopic compounds that can absorb so much moisture from their surroundings that they eventually dissolve in the absorbed water. An example of this type of compound is calcium chloride and called as deliquescent salts.

(2). No, taking a new mass would be a combination of anhydrous and hydrated salt. Also the reacted salt would not change the measurement of the temperature during the experiment. As explained in 1. there will be difference in the weight of the salt as compared with the original one, due to absorption of moisture. The student will come to know about the increased weight as he had the previous original weight. The student can avoid the error if he is able to identify the reason for the weight change and he may ask for the fresh sample.

(3).The glass beaker would allow more heat energy to escape from the solution than the pressed polystyrene cup would have done. If heat is released into the room through the container, the measurement for enthalpy won't be accurate. Better is to use the insulated,polystrene cup.

2.H = Cp x T

H = Cp x (T2 – T1)

So, if thermometer shows the BP of water at 103.5C and freezing point of ice at 3.5 C , the value of T will be 100C which will be same as if thermometer marks BP at 100C and FP at 0C. Hence as shown above H depends on the value of T, which do not change so, it don’t show any impact on the H of the experiment.

But this error has to be taken under consideration if same thermometer is being used to take temperature of other substance.

3.

G = H - TS, where H is the enthalpy change and S is the entropy change.

Because the crystal has much less entropy (disorder) than an amorphous (unstructured or chaotic) solid, S is negative. That means that crystallization of hydrated salt must be "enthalpy driven," and it is. H is negative and hydrates are usually more thermodynamically stable then anhydrous salts, so they form spontaneously takes atmospheric humidity. Water of hydration helps hydrated salt to take up well ordered crystals., Hence hydrated salts have high crystal lattice energy to break the regular fashion in comparision to anhydrous salt. The salvation energy of hydrated salt is more.

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