Question

(a) The mass density of a gaseous compound was found to be 1.23 kg m^−3 at 330 K and 20 kPa. What is the molar mass of the compound? (b) In an experiment to measure the molar mass of a gas, 250 cm3 of the gas was confined in a glass vessel. The pressure was 152 Torr at 298 K, and after correcting for buoyancy effects, the mass of the gas was 33.5 mg. What is the molar mass of the gas?

Answer 1

Answer:

The molar mass of the compound is:- 168.82 g/mol

The molar mass of the gas is:- 16.38 g/mol

Explanation:

(a)

Using ideal gas equation as:

$PV=nRT$

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Also,  

Moles = mass (m) / Molar mass (M)

Density (d)  = Mass (m) / Volume (V)

So, the ideal gas equation can be written as:

$PM=dRt$

Given that:-

Pressure = 20 kPa = 20000 Pa

The expression for the conversion of pressure in Pascal to pressure in atm is shown below:

P (Pa) = $\frac {1}{101325}$ P (atm)

20000 Pa = $\frac {20000}{101325}$ atm

Pressure = 0.1974 atm

Temperature = 330 K

d = 1.23 kg/m³ = 1.23 g/L

Molar mass = ?

Applying the equation as:

0.1974 atm × M = 1.23 g/L × 0.0821 L.atm/K.mol × 330 K

⇒M = 168.82 g/mol

The molar mass of the compound is:- 168.82 g/mol

(b)

Given that:

Pressure = 152 Torr

Temperature = 298 K

Volume = 250 cm³ = 0.25 L

Using ideal gas equation as:

$PV=nRT$

R = $62.3637\text{torr}mol^{-1}K^{-1}$

Applying the equation as:

152 Torr × 0.25 L = n × 62.3637 L.torr/K.mol × 298 K

⇒n = 0.002045 moles

Given that :  

Mass of the gas = 33.5 mg = 0.0335 g

Molar mass = ?

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$0.002045\ moles = \frac{0.0335\ g}{Molar\ mass}$

The molar mass of the gas is:- 16.38 g/mol