Question

QUESTION 1 is 1.34 g/L at STP The density of {Hint: these molecules have different molar masses.} NO co Ne F2 CH4 QUESTION 2 The density (in g/L) of CO2 at 1157.5 torr and 65.3 °C is 12.5 0,0238 0.0548 1830 2.41 QUESTION 3 g/L The density of nitric oxide (NO) gas at 0.866 atm and 29.8 °C is 1.05 0.354 0.957 10.6 0,0348

Answer 1

1)

P = 1.0atm

T = 273 K

density = 1.34 g/L

Lets derive the equation to be used

use:

p*V=n*R*T

p*V=(mass/molar mass)*R*T

p*molar mass=(mass/V)*R*T

p*molar mass=density*R*T

Put Values:

1 atm * MM = 1.34g/L * 0.08206 atm.L/mol.K *273.0 K

MM = 30.0 g/mol

This is molar mass NO

Answer: NO

2)

P= 1157.5 torr

= (1157.5/760) atm

= 1.523 atm

T= 65.3 oC

= (65.3+273) K

= 338.3 K

Molar mass of CO2,

MM = 1*MM(C) + 2*MM(O)

= 1*12.01 + 2*16.0

= 44.01 g/mol

Lets derive the equation to be used

use:

p*V=n*R*T

p*V=(mass/molar mass)*R*T

p*molar mass=(mass/V)*R*T

p*molar mass=density*R*T

Put Values:

1.523 atm *44.01 g/mol = density * 0.08206 atm.L/mol.K *338.3 K

density = 2.4145 g/L

Answer: 2.41

3)

P = 0.866atm

T= 29.8 oC

= (29.8+273) K

= 302.8 K

Molar mass of NO,

MM = 1*MM(N) + 1*MM(O)

= 1*14.01 + 1*16.0

= 30.01 g/mol

Lets derive the equation to be used

use:

p*V=n*R*T

p*V=(mass/molar mass)*R*T

p*molar mass=(mass/V)*R*T

p*molar mass=density*R*T

Put Values:

0.866 atm *30.01 g/mol = density * 0.08206 atm.L/mol.K *302.8 K

density = 1.0459 g/L

Answer: 1.05