Question

The mass of a gas that occupies a volume of 612.5 ml at room temperature and pressure (25° C and 1 atm pressure) is 1.1g. The molar mass of the gas is
(a) 66.25 g mol⁻¹
(b) 44 g mol⁻¹
(c) 24.5 g mol⁻¹
(d) 662.5 g mol⁻¹

Answer 1

We convert the volume to the volume standard condition (STP)
V1 = 612.5 ml
P1 = 1 atm
T1 = 25 + 273 = 298K
V2 = ?
P2 = 1 atm
T2 = 273


Since P1 =P2, we use the formula

V1/T1 = V2/T2

612.5 / 298 = V2/273

V2 = (612.5x273) / 298 = 561.1 ml = 0.561 Litres


Molar gas volume at STP = 22.4 litres

Number of moles of the gas = 0.561/22.4 = 0.025 moles

Molar mass = Given mass/Number of moles

Molar mass = 1.1/0.025 
                   = 44g/mol
Choice (b)

Answer 2

Answer:

Explanation:

We convert the volume to the volume standard condition (STP)

V1 = 612.5 ml

P1 = 1 atm

T1 = 25 + 273 = 298K

V2 = ?

P2 = 1 atm

T2 = 273

Since P1 =P2, we use the formula

V1/T1 = V2/T2

612.5 / 298 = V2/273

V2 = (612.5x273) / 298 = 561.1 ml = 0.561 Litres

Molar gas volume at STP = 22.4 litres

Number of moles of the gas = 0.561/22.4 = 0.025 moles

Molar mass = Given mass/Number of moles

Molar mass = 1.1/0.025

= 44g/mol

Choice (b)