Question

Q11.It was found that 380mL of a gas at 27◦C and 800mm of Hg weighed 0.455g. Find the molecular weight of the gas.

Answer 1

The standard way to do this problem is to use the gas law equation:
PV = nRT
P = pressure = 800/760 = 1.053 atm
V = volume = 0.380L
n = number of moles
R = gas law constant = 0.082057
T = temp in K = 27+273 = 300
Substitute:
1.053* 0.380 = n * 0.082057*300
n = ( 1.053*0.380) / ( 0.082057*300)
n = 0.400/ 24.6171
n = 0.0162 mol

0.455g = 0.0162 mol
1mol = 0.455/ 0.0162 = 28.00g/mol

Answer 2

Given conditions ⇒

Pressure of the Gas(P) = 800 mm of Hg.
= 800/760 atmospheric pressure.
= 1.052 atm.

Volume of the gas(V) = 380 ml.
 = 0.380 liter.
Temperature at which the gas is kept(T) = 27°C
= 273 + 27
= 300 K.

Now, Using the Formula, 
PV = nRT
Where, n = number of moles of the gas,
R is the gas constant = 8.2 × 10⁻² liter atm mole⁻¹ K⁻¹

Now, 
1.052 × 0.380 = n × 8.2 × 10⁻² × 300
∴ n = 0.39976/24.6
∴ n = 0.0162

Now, Given mass of the gas = 0.455 grams.

∴ Using the Formula,
  No. of the moles = Mass/Molar Mass.
 ∴ Molar Mass = Mass/No. of moles.
 ∴ Molar Mass = 0.455/0.0162
⇒ Molar Mass = 28.08 g/moles.


Hence, the molar mass or the molecular weight of the gas is 28.08 g/mole.


Hope it helps.