The molar mass of the vapor of trial 1 and trial 2 will be 54.68 g/mol and 57.68 g/mol respectively.
For Trial 1:
(Following values are missing in question)
T = 101 °C or 374.15 K
P = 760 mmHg or 1 atm
V = 292 ml or 0.92 L
R= O.08206 L.atm.K⁻¹.mol⁻¹
mass of vapor (m) = 0.52 g
Calculate the molar mass of vapor by using the ideal gas law
PV =(m/M) RT
Rearrange it for molar mass
M = mRT / PV
M = 0.52 g × 0.08206 L.atm.K⁻¹.mol⁻¹ × 374.15 K / 1 atm × 0.92 L
M = 54.68 g/mol
For Trial 2:
(Following values are missing in question)
T = 373.15 K
P = 1 atm
V = 0.292 L
R= O.08206 L.atm.K⁻¹.mol⁻¹
mass of vapor (m) = 0.55 g
Calculate the molar mass of vapor by using the ideal gas law
PV =(m/M) RT
Rearrange it for molar mass
M = mRT / PV
M = 0.55 g × 0.08206 L.atm.K⁻¹.mol⁻¹ × 373.15 K / 1 atm × 0.292 L
M = 57.68 g/mol
You can also learn about ideal gas law from the following question:
https://brainly.com/question/28257995
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