Answer 1 of 1 Done (1)mass of flask, aluminum foil, and rubber band, 80.772g (2)
ID: 702277 • Letter: A
Question
Answer 1 of 1 Done (1)mass of flask, aluminum foil, and rubber band, 80.772g (2) temperature of boiling water, 100 °C = 100 + 273 15 K (3) Atmosphernic pressure, P- 101.325 kPa-1 atnm we know 1 atm 101325 kPa 373 15 K (4) mass of flask, aluminum foi·rubber band, and condensed water vapor, m-81 959 g mass of condensed vapor m-m-81.959 g-80.772 g 1.187 g mass of flask, aluminum foil, rubber band, and water, m210 694g - = 210 694 g-80 772 g = 129 922 g volume of water (8) mass of water = (9)volume of flask mass of water density of water 129.922 g = 129, 922mL-IL -0.129922L (10)volume of gas volume of flask 1000 mL V= 0.129922 L PV MRT PT RT 1 atm 0 129922 L (0.08206 L atm/mol K)x373.15K 0.004243 mol mass of gas (12)molar massof gmos of ga 1.187 g 279 76g/mol ;Explanation / Answer
Ans 1
Mass of condensed vapor = m2 - m1
= 81.959 - 80.772
= 1.187 g
Ans 2
Mass of water in flask = m3 - m1 = 210.694 - 80.772
= 129.922 g
Density of water = 1 g/mL
Volume of test tube = mass/density
= 129.922 / 1g/ml
= 129.922 mL = 0.129922 L
Ans 3
From the ideal gas equation
PV = nRT
Moles n = PV/RT
n = 1 atm x 0.129922 L / 0.08206 L-atm/mol-K x 373.15 K
n = 0.004243 moles
Molar mass = mass/moles = 1.187 g / 0.004243 moles
= 279.758 g/mol
Ans 4
Unknown liquid = Tribromo(methyl)silane(CH3Br3Si) which has molar mass of 279.758 g/mol