What pressure (in atm) would be exerted by 76 g of fluorine gas (f2) in a 1.50 liter vessel at - 37oc? (a) 26 atm (b) 4.1 atm (c) 19,600 atm (d) 84 (e) 8.2 atm?

Let's assume that the F ₂ gas has ideal gas behavior.

Then we can use ideal gas formula,

PV = nRT

Where, P is the pressure of the gas (Pa), V is the volume of the gas (m³), n is the number of moles of gas (mol), R is the universal gas constant (8.314 J mol ⁻¹ K⁻ ¹) and T is temperature in Kelvin.

Moles = mass / molar mass

Molar mass of F₂ = 38 g/mol

Mass of F₂ = 76 g

Hence, moles of F₂ = 76 g / 38 g/mol = 2 mol

P = ?

V = 1.5 L = 1.5 x 10 ⁻³ m³

n = 2 mol

R = 8.314 J mol⁻¹ K⁻ ¹

T = - 37 °C = 236 K

By substitution,

P x 1.5 x 10⁻³ m³ = 2 mol x 8.314 J mol⁻¹ K⁻¹ x 236 K

p = 2616138.67 Pa

p = 25.8 atm = 26 atm

Hence, the pressure of the gas is 26 atm.

Answer is "a".