# X and Y are two elements which form $X_2Y_3$ and $X_3Y_4$.If 0.20mol of $X_2Y_3$ weighs 32.0g and 0.4mol of $X_3Y_4$ weighs 92.8g.The atomic weights of X and Y are respectively

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

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.