Answer:
The halogen is Iodine.
Explanation:
Using the ideal gas equation, we find the number of moles of gas present, n.
PV = nRT where P = pressure of gas = 1.41 atm, V = volume of gas = 109 mL = 0.109 L, n = number of moles of gas, R = molar gas constant = 0.082 L-atm/mol-K and T = temperature of gas = 398 K
Since PV = nRT, making n subject of the formula, we have
n = PV/RT
substituting the values of the variables into the equation, we have
n = 1.41 atm × 0.109 L/(0.082 L-atm/mol-K × 398 K)
n = 0.15369 atm-L/32.636 L-atm/mol
n = 0.0047 mol
Since n = m/M where m = mass of gas = 1.19 g and M = relative molecular mass of gas
So, M = m/n
M = 1.19 g/0.0047 mol
M = 252.7 g
Since halogens are diatomic the relative atomic mass is M/2 = 252.7g/2 = 126.34 g
From tables, the only halogen with this atomic mass is Iodine.
So, the halogen is Iodine.
