Question

Calculate to three significant digits the density of boron trifluoride gas at exactly 20 C and exactly 1atm . You can assume boron trifluoride gas behaves as an ideal gas under these conditions.

Answer 1

Answer:

0.5583  g/L[/tex]

Explanation:

Since boron trifluoride ( B$F_{3}$ ) 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 $JK^{-1} mol^{-1}$

T - temperature of the gas ( 273 + 20 = 298 K )

thus ,

$20*V = 1*8.314*293\\V= 121.8001 L$

density ρ = $\frac{mass}{volume}$

mass of  B$F_{3}$ is :

B : 11

F : 19

therefore , mass = 11 + $3*19$

=68 g

density = $\frac{68}{121.8001} = 0.5583 g/L$