A sample of an unknown volatile liquid was injected into a Dumas flask ( mflask =27.0928g. Vflask=0.104L) and heated until no visible traces of the liquid could be found. The flask and its contents were then rapidly cooled and reweighed (mflask+vapor=27.4593g) . The atmospheric pressure and temperature during the experiment were 0.976 atm and 18.0 °C, respectively. The unknown volatile liquid was ________
Answer:
The unknown volatile liquid that has an approximate value of a molar mass of 86.23 g/mol is Hexane ( C₆H₁₄)
Explanation:
From the above given question:
the mass of the liquid = mflask+vapor - mflask
the mass of the liquid = 27.4593g - 27.0928g = 0.3665 g
The volume of the Dumas Flask = 0.104 L
The pressure of the Dumas Flask = 0.976 atm
The temperature of the Dumas Flask = 18.0°C = (18 + 273) = 291 K
Using Ideal gas equation:
PV = nRT
0.976 atm × 0.104 L = n × 0.08205 L atm /mol/K × 291 K
n = (0.976 atm × 0.104 L )/ ( 0.08205 L atm /mol/K × 291 K)
n = ( 0.101504 atm L)/ 23.87655 L atm /mol)
n = 0.004251 mol
We know that ;
number of moles = mass/molar mass
∴
0.004251 mol = 0.3665 g/molar mass
molar mass = 0.3665 g / 0.004251 mol
molar mass = 86.23 g/mol
The unknown volatile liquid that has an approximate value of a molar mass of 86.23 g/mol is Hexane ( C₆H₁₄)
