A sample of an unknown volatile liquid was injected into a Dumas flask and heated until no visible traces of the liquid could be found. The flask and its contents were then rapidly cooled and reweighed The atmospheric pressure and temperature during the experiment were 0.976 atm and 18.0 °C, respectively. The unknown volatile liquid was

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

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)