The following two question are from the molar mass of a volatile liquid lab and

Question

The following two question are from the molar mass of a volatile liquid lab and the specific heat of our "pet stones" lat. Molar mass lab. Remember to calculate the molar mass of a volatile liquid the equation that we used was: Pmm = gRTN, where P is the barometric pressure in atm.; mm is the molar mass of the volatile liquid; g is the grams of vapor; R is the ideal gas constant, and T is the temperature in Kelvin. Following you will be presented with some procedural errors that were carried out by you and your partner(s). You task is to determine how these mentioned procedural error will affect the results of the experiment. a). You have your lab partner get the mass of the flask and stopper and as these things go your partner is a digit inverter. The balance reads that the mass of the flask and stopper is 256.132 grams, your partner reports the mass as 265.132 grams. b). In your haste to get out the door, after all it is a sunny day in November, you forget to convert the barometric pressure, recorded as mm Hg, to atm. c). You forget, or maybe it was your partner, to convert the vapor temperature to Kelvin. To get full credit, after all you have a 1 in 3 change of guessing the right answer, you must supply some quantitative back up for your answer. Specific heat lab: What is the principlels) used in deterring the specific heat of your pet stone and how is that principle(s) actually utilized in the doing the experiment? 8.

Explanation / Answer

a) Suppose the mass of the flask, stopper and the vapor is 266.000 g (hypothetical). The mass of empty flask and stopper is actually 256.132 g; however, due to the error, the mass of the empty flaks and stopper was recorded as 265.132 g. The correct mass of the vapor should have been (266.000 – 256.132) g = 9.868 g; however, due to the error, the mass of the vapor was incorrectly obtained as (266.000 – 265.132) g = 0.868 g. Offcourse, the mass of the vapor was calculated to be much lower than the actual value. Since the mass of the vapor (g) is directly proportional to the molar mass of the vapor (mm), hence, the calculated molar mass of the vapor will be much lower than the actual value.

b) The procedure mentions clearly that the pressure of the vapor (P) needs to be recorded in atmospheres. However, the pressure was actually recorded in mmHg. Suppose the pressure was read as 760 mmHg (hypothetical again). We know that 760 mmHg = 1 atm; however, since you didn’t convert the pressure in mmHg to atm, you will put down the pressure as 760 (numeric value) in your calculations. We note that P is inversely proportional to mm as per the given expression; the higher the pressure recorded, the lower will be mm. Consequently, with such a high value of P (in mmHg), the calculated mm will be much lower than the actual value.

c) The procedure asks us to record the vapor temperature in the Kelvin scale. Suppose the temperature was carried out at room temperature, 25°C. The absolute temperature is (25 + 273) K = 298 K. However, you noted the temperature as 25 (numeric value). We note from the expression that mm is directly proportional to T. Since T was recorded low (25 vs 298), the calculated molar mass, mm of the vapor will be calculated low.

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