The given figure shows the volume ( V ) versus absolute temperature...

According to the ideal gas equation, $PV = nRT$.

Substituting the number of moles $n = \dfrac{m}{M}$, where $m$ is the mass and $M$ is the molar mass, we get:

$PV = \dfrac{m}{M}RT$

Rearranging this equation to represent the volume $V$ as a function of absolute temperature $T$, we have:

$V = \left(\dfrac{mR}{PM}\right)T$

This represents a straight line passing through the origin with a slope given by:

$\text{Slope} = \dfrac{mR}{PM}$

Since the mass $m$ and pressure $P$ are constant for all three gases, the slope is inversely proportional to the molar mass $M$:

$\text{Slope} \propto \dfrac{1}{M}$

From the given $V-T$ graph, the slopes of the lines are in the order:

$\text{Slope}_A > \text{Slope}_B > \text{Slope}_C$

Therefore, the relation between their molar masses is:

$M_A < M_B < M_C$

Answer: $M_A < M_B < M_C$