Elements A and B form two compounds B_2A_3 an | vedclass

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(C) Let the atomic weights of $A$ and $B$ be $a$ and $b$ respectively.For $B_2A_3$: $0.05 	ext{ moles} = 9.0 \ g$,so $1 	ext{ mole} = \frac{9.0}{0.0

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$40$ and $30$

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(C) Let the atomic weights of $A$ and $B$ be $a$ and $b$ respectively.For $B_2A_3$: $0.05 	ext{ moles} = 9.0 \ g$,so $1 	ext{ mole} = \frac{9.0}{0.05} = 180 \ g/mol$.Thus,$2b + 3a = 180 \dots (i)$For $B_2A$: $0.10 	ext{ mole} = 10 \ g$,so $1 	ext{ mole} = \frac{10}{0.10} = 100 \ g/mol$.Thus,$2b + a = 100 \dots (ii)$Subtracting equation $(ii)$ from $(i)$:$(2b + 3a) - (2b + a) = 180 - 100$$2a = 80 \implies a = 40$.Substituting $a = 40$ in equation $(ii)$:$2b + 40 = 100$$2b = 60 \implies b = 30$.Therefore,the atomic weights of $A$ and $B$ are $40$ and $30$ respectively.

Elements $A$ and $B$ form two compounds $B_2A_3$ and $B_2A$. $0.05$ moles of $B_2A_3$ weigh $9.0 \ g$ and $0.10$ mole of $B_2A$ weigh $10 \ g$. The atomic weights of $A$ and $B$ are:

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