$\begin{array}{1 1}(a)\;\text{16.0 and 56.0}\\(b)\;\text{8.0 and 28.0} \\(c)\;\text{56.0 and 16.0}\\(d)\;\text{28.0 and 8.0}\end{array} $

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Molecular weight of $X_2Y_3=\large\frac{32.0g}{0.2mol}$$=160gmol^{-1}$

Molecular weight of $X_3Y_4=\large\frac{92.8g}{0.4mol}$$=232gmol^{-1}$

Now suppose the atomic weight of X=a and atomic weight of $Y=b$

Then $2a+b=160$ and $3a+4b=232$

Solving the equations we get,

$b=16$ and $a=56$

Hence (c) is the correct answer.

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