Question

0.10 g of the volatile liquid X formed 0.025 dm3 of vapour at 100°C and atmospheric pressure.
1 mol of vapour occupies 22.4 dm3 at 0°C and atmospheric pressure.
What is the relative molecular mass of X?
A (0.025 x 273x22.4)/
(0.10 x 373)
B (0.025 x 273x22.4)/
(0.10 x 273) C (0.10 x 273x22.4)/
(0.025x373) D (0.10 x 373x22.4)/
(0.025x273)

Answer 1

Answer:

0.1g = 0.000817 mol.

Explanation:

Disclaimer:

I am not used to doing this sort of calculations, and am not sure if the ideal gas law can be used in this way.

My answer is dependant on a few assumptions.

Volatile liquid X in its gaseous state is an ideal gas

Step 1: Manipulation of ideal gas law

Using the ideal gas law:

PV = nRT

P/R = nT/V

Step 2: Substituting the numbers in the question

0.10g of the volatile liquid X formed 0.025dm3 of vapour at 100°C and atmospheric pressure. (First statement) 1 mol of vapour occupies 22.4dm3 at 0°C and atmospheric pressure. (Second statement)

The Celsius Scale cannot be used as it does not measure absolute temperature. As such, the Kelvin Scale has to be used.

0 deg.C = 273.15K

100 deg.C - 373.15K

P/R = n(373.15K)/0.025dm3 {First statement]

P/R = 1(273.15K)/22.4dm3 [Second statement]

The P/R for both the first situation (0.1g at 100 deg. C) and the second situation, (1 mol. at 0 deg. C) are equal, as both pressures are at 1 atm, and R is a constant.

n(373.15K)/0.025dm3 = 1(273.15)/22.4dm3

Step 3: Solving for n

n = [(273.15)/22.4]/[(373.15)/0.025] = 0.000817mol.