When 0.32 g of sulfur is burnt in air,224 m | vedclass

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(A) The molar mass of $SO_2$ is $32 + 32 = 64 \ g/mol$. At $N.T.P$,$22400 \ mL$ of $SO_2$ weighs $64 \ g$. Therefore,$224 \ mL$ of $SO_2$ weighs $\fra

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Definite Proportions

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(A) The molar mass of $SO_2$ is $32 + 32 = 64 \ g/mol$. At $N.T.P$,$22400 \ mL$ of $SO_2$ weighs $64 \ g$. Therefore,$224 \ mL$ of $SO_2$ weighs $\frac{224 	imes 64}{22400} = 0.64 \ g$. In the first experiment,$0.64 \ g$ of $SO_2$ contains $0.32 \ g$ of sulfur. Thus,the percentage of sulfur is $\frac{0.32}{0.64} 	imes 100 = 50\%$. In the second experiment,the compound also contains $50\%$ sulfur. Since the composition of sulfur and oxygen in $SO_2$ remains constant regardless of the source or method of preparation,it follows the Law of Definite Proportions.

Compound	$\% Sn$	$\% O$
$A$	$78.77$	$21.23$
$B$	$88.12$	$11.88$

When $0.32 \ g$ of sulfur is burnt in air,$224 \ mL$ of $SO_2$ is obtained at $N.T.P$. In another experiment,$SO_2$ is formed by the decomposition of a compound containing $50\%$ sulfur. The results of the given information follow the law of ......

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Compound $\% Sn$ $\% O$

$A$ $78.77$ $21.23$

$B$ $88.12$ $11.88$

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