\(\because 0.1 \mathrm{mole}\)
\(X Y_{2} \equiv 10 g \therefore 1 \mathrm{mole}\)
\(X Y_{2} \equiv 100 g=X+2 Y=100 \ldots (1)\)
For \(X_{3} Y_{2}\)
\(\because 0.05\) mole
\(X_{3} Y_{2}\)
\(\because 9 g\)
\(\therefore 1\) mole \(X_{3} Y_{2} \equiv 180 g\)
\(3X+2Y=180\dots (2)\)
On solving \((1)\) and \((2)\), \(X=40\) And \(Y=30\)
(આપેલ છે: આણ્વિય દળ ${C}=12, {H}=1, {O}=16 {u}$ )