Question

At 273K the density of a gaseous oxide at 2 bar is the same as that of nitrogen at 5 bar. Calculate the molecular mass of oxide.

Answer 1

Hint: The ideal gas law has to be used in this question. The ideal gases are those which have the tendency to produce elastic collisions between the molecules and do not have the intermolecular forces of attraction. These ideal gases molecules have random motion and are rigid spheres.

Complete step by step answer:
The gases which obey the Charle’s law, Boyle’s law and the Avogadro law are the ideal gases. So the formula for ideal gases is the following:
pV=nRT
p=pressure
V=volume
n=moles
R= constant
T= temperature
So the moles is the given mass of the solute by its molar mass so we can substitute the moles in the above formula by the definition I gave. So I will write it as:
$pV = \dfrac{w}{M}RT$
W= mass given
M= molar mass
On interchanging the places of M by Vwe get,
$pM = \dfrac{w}{V}RT$
So the density defination is the mass by volume so in the formula we are getting at formula so we will replace it by d density.
$pM = dRT$
So the R and T are constant here and from the question we came to know that the densities for the gaseous oxide and nitrogen is same that is
${d_1} = {d_2}$
So we get,
${p_1}{M_1} = {p_2}{M_2}$
On substituting the values we get,
$2 \times {M_1} = 5 \times 32$
On further simplifying we get,
$M_1 = \dfrac{5\times 32}{{2}}$
$\therefore M_1 =80 gmol^{-1}$

So the molar mass is $80gmo{l^{ - 1}}$.

Note: The Charle’s law state that at the constant number of moles and the pressure the volume of the gas given is directly proportional to the temperature. the Boyl’s law state that the volume is inversely proportional to the pressure on constant number of moles and temperature. the Avogadro law state that volume is directly proportional to number of moles on constant pressure and temperature.