estimate the density of air (g/L) at 40 degrees celsius and 3 atm. Report answer in units of g/L & two significant figures.

2) Estimate the molecular weight of Linanyl acetate (C12 H20 O2) in terms of (g/mol) using 3 significant figures.

Respuesta :

Answer:

1) The density of air at 40 degrees Celsius and 3 atm pressure is 3.4 g/L.

2) Molecular mass of linanyl acetate  is 196 g/mol.

Explanation:

1) Average molecular weight of an air ,M= 28.97 g/mol

[tex]PV=nRT[/tex]

or [tex] PM=dRT[/tex]

P = Pressure of the gas

T = Temperature of the gas

d = Density of the gas

M = molar mass of the gas

R = universal gas constant

P = 3 atm, T = 40°C = 313.15 K, M = 28.97 g/mol

[tex]d=\frac{PM}{RT}=\frac{3 atm \times 28.97 g/mol}{0.0821 atm L/ mol K\times 313.15 K}[/tex]

d = 3.4 g/mL

The density of air at 40 degrees Celsius and 3 atm pressure is 3.4 g/L.

2) Molecular formula of Linanyl acetate = [tex]C_{12}H_{20}O_2[/tex]

Atomic mass sof carbon  = 12.01 g/mol

Atomic mass of hydrogen = 1.01 g/mol

Atomic mass of oxygen  = 16.00 g/mol

Molecular mass of  Linanyl acetate :

[tex]12\times 12.01 g/mol+20\times 1.01 g/mol+2\times 16.00 g/mol =196.32 g/mol \approx 196 g/mol[/tex]