The ratio of masses of oxygen and nitrogen in a particular gaseous mixture is 1 : 4. The ratio of number of their molecules is:

We are told that in the gaseous mixture the mass of oxygen and the mass of nitrogen are in the ratio $$1:4$$. We wish to obtain the ratio of the number of molecules (which is the same as the ratio of the number of moles) of the two gases.

First, recall the fundamental relation that connects mass, molar mass and number of moles:

$$\text{Number of moles} \; (n)=\dfrac{\text{Given mass}\; (m)}{\text{Molar mass}\; (M)}$$

For a diatomic molecule such as oxygen, $$O_2$$, the molar mass is obtained by doubling the atomic mass of oxygen:

$$M(O_2)=2\times16\;\text{g mol}^{-1}=32\;\text{g mol}^{-1}$$

Similarly, for diatomic nitrogen, $$N_2$$, we have:

$$M(N_2)=2\times14\;\text{g mol}^{-1}=28\;\text{g mol}^{-1}$$

According to the statement of the problem, let the actual masses present be represented by $$m_O$$ and $$m_N$$ such that

$$m_O:m_N=1:4$$

We may now assign convenient proportional masses that satisfy this ratio. An easy choice is

$$m_O=1\,\text{unit},\quad m_N=4\,\text{units}$$

Using the formula for the number of moles, we determine:

For oxygen,

$$n_O=\dfrac{m_O}{M(O_2)}=\dfrac{1}{32}$$

For nitrogen,

$$n_N=\dfrac{m_N}{M(N_2)}=\dfrac{4}{28}=\dfrac{1}{7}$$

We now form the ratio of the numbers of molecules (or moles):

$$n_O:n_N=\dfrac{1}{32}:\dfrac{1}{7}$$

To simplify such a ratio, we multiply both terms by the common denominator $$32\times7$$ so that no fractions remain:

$$n_O:n_N=\left(\dfrac{1}{32}\right)\times(32\times7):\left(\dfrac{1}{7}\right)\times(32\times7)$$

$$\quad\;=7:32$$

Thus the required ratio of the number of molecules of $$O_2$$ to $$N_2$$ is $$7:32$$.

Hence, the correct answer is Option B.