What is the approximate volume of 19g of fluorine gas at a pressure of 4.0 atmospheres and a temperature of 127o c?

V = 8.21 L

Explanation:

Given data:

Volume of fluorine = ?

Mass of fluorine = 19 g

Pressure = 4.0 atm

Temperature = 127 °C(127+ 273 =400 K)

Solution:

Number of moles of fluorine:

Number of moles = mass/molar mass

Number of moles = 19 g/ 19 g/mol

Number of moles = 1 mol

PV = nRT

V = nRT/P

V = 1 mol ×0.0821 atm. L. mol⁻¹. k⁻¹ × 400 K/ 4 atm

V = 32.84 L/ 4

V = 8.21 L

for points guys