Given,$50mL$of a gas A diffuses through a membrane in the same time as for the diffusion of$40mL$ a gas B under identical pressure temperature conditions. If the molecular weight of $A = 64$ then the molecular weight of B would be:
A.$100$
B.$250$
C.$200$
D.$80$

Hint: We have to know that Graham's law of emission was planned by Scottish actual scientific expert Thomas Graham in $1848$ . Graham discovered tentatively that the pace of emission of a gas is contrarily corresponding to the square foundation of the molar mass of its particles.

Complete answer:
We have to see, Graham's law expresses that the pace of dissemination or of emission of gas is conversely corresponding to the square foundation of its sub-atomic weight. In this manner, if the atomic load of one gas is multiple times that of another, it would diffuse through a permeable fitting or break through a little pinhole in a vessel at a large portion of the pace of the other. A total hypothetical clarification of Graham's law was given years after the fact by the dynamic hypothesis of gases. Graham's law gives a premise to isolating isotopes by dispersion, a technique that came to assume a significant part in the improvement of the nuclear bomb.
By using the following formula,
$\dfrac{{Rate - A}}{{Rate - B}} = \sqrt {\dfrac{{{M_2}}}{{{M_1}}}} $
Where,
$Rate - A$= $50mL$
$Rate - B$= $40mL$
${M_1}$= $64g$
${M_2}$has to be calculated.
Applying all the values in the above equation,
$\dfrac{{50mL}}{{40mL}} = \sqrt {\dfrac{{64g}}{{{M_2}}}} $
Rewrite the above expression,
$\sqrt {{M_2}} = \dfrac{5}{4} \times 8 = 10$
Then, ${M_2} = {10^2}$
Hence, ${M_2} = 100g$

Therefore, the correct option is (A).

Note:
We have to know that gases with various densities can be isolated utilizing Graham's law. It is likewise useful in deciding the molar mass of obscure gases by contrasting the pace of dissemination of obscure gas to known gas. We can isolate the isotopes of a component utilizing Graham's law.