Answer:
0.5583 g/L[/tex]
Explanation:
Since boron trifluoride ( B[tex]F_{3}[/tex] ) Is an ideal gas , we can apply IDEAL GAS EQUATION which is ,
PV = nRT
Where ,
P - the pressure at which it is present (20 atm)
V - volume of the gas (needed)
n - number of moles of the gas taken (1 mol)
R - universal gas constant which is 8.314 [tex]JK^{-1} mol^{-1}[/tex]
T - temperature of the gas ( 273 + 20 = 298 K )
thus ,
[tex]20*V = 1*8.314*293\\V= 121.8001 L[/tex]
density ρ = [tex]\frac{mass}{volume}[/tex]
mass of B[tex]F_{3}[/tex] is :
B : 11
F : 19
therefore , mass = 11 + [tex]3*19[/tex]
=68 g
density = [tex]\frac{68}{121.8001} = 0.5583 g/L[/tex]
