GC_4253_L14_Molar_Vapor_Density

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Texas Tech University *

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1117

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Chemistry

Date

Dec 6, 2023

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docx

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6

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PRE-LAB QUESTIONS 1. How would leaving molecular sieves, (which are used as desiccants) uncapped or in an open beaker for an extended period of time affect their functionality? A desiccant is a substance which is hygroscopic in nature, that means it absorbs water. And due to this property, it is used as a drying agent. But if it is left open or uncapped for an extended period of time then it will absorb water from the air, and its ability to absorb water from our required substance would decrease. It functionality will decrease. 2. The barometric pressure is 752.3 mm Hg. What is the pressure in atmospheres? P = 752.3 mm Hg 760 mm Hg = 1 atm 752.3 mm Hg = 1 760 × 752.3 = 0.98 atm 3. Convert 95.8 °C to Kelvin. T = 95.8 °C In K = 273 + °C => 273 + 95.8 => 368.8 K 4. A 3.28 g sample of gas occupies 1.25 L at standard temperature pressure (STP; 0 °C and 1 atm) conditions. What is its molar mass? Using ideal gas law, PV = nRT P = 1 atm, V =1.25 L, R = 0.082 L · atm/mol · K , T = 273 K Molar Mass and Vapor Density
n = Pv RT = 1 × 1.25 0.082 × 273 = 5.58 x 10 2 mol n = given mass molarmass => molar mass = 3.28 g 5.58 × 10 12 = 58.78 g mol molar mass = 58.78 g mol 5. Given the following data, calculate the molar mass. Barometric pressure: 753.1 mm Hg Mass of the flask and foil: 80.268 g Mass of the flask, foil, and condensed liquid: 80.615 g Volume of flask: 137.5 mL Temperature of water bath: 99.4 °C P = 753.1 mm Hg = 753.1/760 = 0.991 atm V = 137.5 ml = 0.1375 L T = 372.4 K R = 0.082 atm. L . mol 1 . K 1 PV = nRT => n = PV/RT => 0.991 × 0.1375 0.082 × 372.4 = 4.46 x 10 3 mol Mass of Liquid = 80.615 – 80.268 = 0.347 g n = 0.347 m = 0.347 4.46 × 10 3 = 77.8 g Molar Mass and Vapor Density
Molar mass = 77.8 g mol Molar Mass and Vapor Density
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