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.

Answer 1

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

$PV=nRT$

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

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

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 :

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