A mixture of gases contains $\mathop H\nolimits_2 $ and $\mathop O\nolimits_2$ gases in the ratio of 1:4 (w/w). What is the molar ratio of the two gases in the mixture?
(a) $16:1$
(b) $2:1$
(c) $1:4$
(d) $4:1$

Hint: It is known to us that the molecular weight of oxygen is 16 amu and the molecular weight of is 2 amu.

Formula: Formula to calculate number of moles is expressed as $n = \dfrac{w}{M}$ ,where is the number of moles, w is the mass of given substance and M is the molecular weight of given substance.

Complete step by step solution:
Try to recall that mole is the amount of any substance. It can be expressed as grams, liters, atoms, molecules or particles.
For any substance, if weight of the substance is given and we have to find out the number of moles of given substance then we can simply find it out by just dividing the weight of given substance with molecular weight of that substance.
Now coming to the question, where the mass ratio of two gases $\mathop O\nolimits_2 $ and $\mathop H\nolimits_2 $ is given 1:4.
Let the mass of $\mathop H\nolimits_2 $ be $x$.
Then, the mass of $\mathop O\nolimits_2 $ will be $4x$ in order to maintain the ratio 1:4.
Also, molecular weight of $\mathop H\nolimits_2 $=2 amu
And the molecular weight of $\mathop O\nolimits_2 $=32 amu.

So, by applying the above formula we can calculate the number of moles for both the gases.
Number of moles of $\mathop H\nolimits_2$ , $\mathop n\nolimits_1 = \dfrac{x}{2}$.
Number of moles of $\mathop O\nolimits_2$ , $\mathop n\nolimits_2 = \dfrac{{4x}}{{32}} = \dfrac{x}{8}$.
In order to find the molar ratio of two gases in the mixture, $\mathop n\nolimits_1 $ divides by $\mathop n\nolimits_2 $.
Hence, $\mathop n\nolimits_1 :\mathop n\nolimits_2 = 4:1$.
Therefore, from above we can say that option A is the correct option.

Note: It should be noted that if number of particles, atoms or molecules of a substance is given then, in that case we can find number of moles just by dividing with Avogadro number which is equal to $6.022×\mathop {10}\nolimits^{23} $.