Two elements $A$ and $B$ form compounds having formula \[A{{B}_{2}}~~\]​ and \[A{{B}_{4}}\] When dissolved in \[20~g~\] of benzene \[\left( {{C}_{6}}{{H}_{6}} \right),1~g~\] of \[A{{B}_{2}}~~\] lowers the freezing point by \[2.3~K~\] whereas \[1.0~g~\] of \[A{{B}_{4}}\] ​ lowers it by \[1.3~K.\]The molar depression constant for benzene is \[5.1~K~kgmo{{l}^{-1}}\]. Calculate atomic masses of $A$ and $B.$

Hint: A German scientist named Dobereiner states that, when elements are arranged into groups of three in the order of their increasing atomic mass, the atomic mass of the element which comes in the middle is the arithmetic mean of the rest of two. On this basis, the three elements in one group are known as ‘Triad’. This arrangement of elements is known as Dobereiner Triads.

Complete step by step answer:
We have compound $A{{B}_{2}}$ :
${{M}_{B}}=\dfrac{{{K}_{f}}\times {{W}_{B}}\times 1000}{{{W}_{A}}\times \Delta {{T}_{f}}}$ and here we have the values for $\Delta {{T}_{f}}=2.3K$ , ${{W}_{B}}=1g$ , ${{K}_{f}}=5.1Kg/mol$
Now after substituting the given values we get the value of ${{M}_{B}}$ as;
${{M}_{B}}=\dfrac{5.1\times 1\times 1000}{20\times 2.3}=110.87g/mol$
Similarly for compound $A{{B}_{4}}$ :
${{M}_{B}}=\dfrac{{{K}_{f}}\times {{W}_{B}}\times 1000}{{{W}_{A}}\times \Delta {{T}_{f}}}$ and here we have the values for $\Delta {{T}_{f}}=1.3K$ , ${{W}_{B}}=1g$ , ${{W}_{A}}=20g$
${{K}_{f}}=5.1Kg/mol$
Now after substituting the given values we get the value of ${{M}_{B}}$ as;
${{M}_{B}}=\dfrac{5.1\times 1\times 1000}{20\times 1.3}=196.154g/mol$
Now, let $a$ \[g/mol\] and $b$ \[g/mol\] the atomic masses of $A$ and $B$ respectively.
\[{{M}_{A{{B}_{2}}}}=a+2b=110.87....(i)\] and
\[{{M}_{A{{B}_{4}}}}=a+4b=196.15....(ii)\]
Subtracting equation (ii) from equation (i), we have
\[-2b=-85.28\]
Atomic mass of $B$ is \[\Rightarrow b=42.64.\]
Substituting the values of b in equation (i), we get,
\[a+2\times 42.64=110.87\]
Atomic mass of $A$ is \[\Rightarrow a=25.59~g/mol.\]
Hence, the atomic mass of $a$ is \[25.59\text{ }g/mol\] and $b$ is \[42.64\text{ }g/mol\]

Note: Other than \[Li,\text{ }Na,\] and \[K,\]elements groups like \[Ca,\text{ }Sr,\] iodine are other examples for Dobereiner Triads. Only a group of three elements in the periodic table comes under Dobereiner Triads. This rule cannot apply for elements such as very low or high atomic mass.