Question

A0.179g sample of an unknown halogen occupies 109 mL at 398 K and 1.41 atm. What is the identity of the halogen? O 12 OF2 O Cl2 O Br2 O Ge

Answer 1

Solution

The given problem can be solved by using the ideal gas equation. Halogens always exist as diatomic gas.
the given gas can be treated as an ideal gas. We can apply the ideal gas equation PV=nRT , where R is ideal gas constant R=0.08212 atmk mol Vearranging the equation we get n - PV RT RT Given data in the question P=1.4latm v=109ml = 109 L = .109 Love 1000 T= 398k n=PV = 1.4 1 atm x 0.109L 0.0821 Latm k mol" x 398K n(no of moles) - 0.15369 = 0.004+ moles 32.6758 Hass of the sample guien = 0.1999 given mass of the sample - molar mass number of moles 0.1799 , 36.0851 g/mol 0.004#mol Halogens excists as diatomic gas 60 38.08519/mol will be the molar mass of a halogen atoms

so dividing the molar mass by two will give the actual identity of the halogen 38:0851g/mol 19.04 g/mol. the final value 19:04 g/mol is almost similian to the molar mass of Fluorine which has molar mass is 998 g/mol Therefore the answer for the guies question is the second option Fa t