When an equal weight mixture of SO_2 and O_2 | vedclass

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(C) Let the weight of both $SO_2$ and $O_2$ be $w \ g$.Molar mass of $SO_2 = 32 + (2 	imes 16) = 64 \ g/mol$.Molar mass of $O_2 = 2 	imes 16 = 32 \

Vedclass · Accepted Answer

$2/3$

Vedclass · Answer

(C) Let the weight of both $SO_2$ and $O_2$ be $w \ g$.Molar mass of $SO_2 = 32 + (2 	imes 16) = 64 \ g/mol$.Molar mass of $O_2 = 2 	imes 16 = 32 \ g/mol$.Number of moles of $SO_2$ $(n_{SO_2})$ = $w/64$.Number of moles of $O_2$ $(n_{O_2})$ = $w/32$.Total moles $(n_{total})$ = $w/64 + w/32 = (w + 2w)/64 = 3w/64$.The partial pressure of a gas is given by its mole fraction multiplied by the total pressure $(P_{O_2} = X_{O_2} 	imes P_{total})$.Mole fraction of $O_2$ $(X_{O_2})$ = $n_{O_2} / n_{total} = (w/32) / (3w/64) = (w/32) 	imes (64/3w) = 2/3$.Therefore,the ratio of the partial pressure of $O_2$ to the total pressure is $2/3$.

When an equal weight mixture of $SO_2$ and $O_2$ is taken in an empty container,the ratio of the partial pressure of $O_2$ to the total pressure is ......

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