Question

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.

Answer 1

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.

Step-by-step explanation:

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

`PV=nRT`

or
`PM=dRT`

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

`d=(PM)/(RT)=(3 atm * 28.97 g/mol)/(0.0821 atm L/ mol K* 313.15 K)`

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 =
`C_(12)H_(20)O_2`

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 :

`12* 12.01 g/mol+20* 1.01 g/mol+2* 16.00 g/mol =196.32 g/mol \approx 196 g/mol`