Question

A mixture of \[{{O}_{2}}\] and gas Y (mol.wt. 80) in the ratio a:b has a mean molecular weight 40. What would be mean molecular weight, if the gases are mixed in the ratio b:a under identical conditions?(gases are non reacting)

Answer 1

Hint: The molecular weight is defined as the weight of a substance in the atomic mass unit of each atom present in the formula. The atomic mass unit is the 1/12 th of the weight of the isotope of carbon-12. It helps us to determine individually how many moles of each atom we need to make that particular compound. 

Complete answer:

Now to approach the question we need to find the molecular weight of each atom present.

The molecular weight of oxygen is = 16(2)= 32g/mol

The molecular weight of Y is = 80g/mol

Mean molecular weight of mixture = 40

Formula for mean molecular weight when the mixture is reacting in ratio of a:b

= \[\dfrac{a(molecular weight of oxygen)+b(molecular weight of Y)}{a+b}\]

Substituting the values in the above formula we get,

$40=\dfrac{a(32)+b(80)}{a+b}$

$40a+40b=32a+80b$

$8a=40b$

$a=5b$ 

Now the formula for mean molecular weight of the mixture when mixture is reacting in ratio of b:a

=\[\dfrac{b(molecular weight of oxygen)+a(molecular weight of Y)}{b+a}\]

Now here we will substitute a=5b and substituting other values we get,

Mean molecular weight=\[\dfrac{b(32)+5b(80)}{b+5b}\]

Mean molecular weight= \[\dfrac{32b+400b}{b+5b}=\dfrac{432b}{6b}=72\]

So, the mean molecular weight is 72.

Note: The 1 mole of any substance is equal to \[6.023\times {{10}^{23}}\] molecules. The mass of one mole of carbon 12 atoms is exactly equal to 12 grams and also its  molar mass is equal to 12 grams per mole. We can also find the composition percentage of the substance by just dividing the mass of the given substance by the total mass of the substance