A mixture of SO_(2) and O_(2) in the molar ratio 16:1 is diffused through a pin hole for successive effusion three times to give a molar ratio 1:1 of diffused mixture. Which one are not correct if diffusion is made at same P and T in each operation? (I) Eight operation are needed to get 1:1 molar ratio. (II) Rate of diffusion for SO_(2):O_(3) after eight operations in 0.707. (III) Six operations are needed to get 2:1 molar ratio for SO_(2) and O_(2) in diffusion mixture. (IV) Rate of diffusion for SO_(2) and O_(2) after six operations is 2.41.

A mixture of SO_(2) and O_(2) in the molar ratio 16:1 is diffused thro

1.	A mixture of SO_(2) and O_(2) in the molar ratio 16:1 is diffused through a pin hole for successive effusion three times to give a molar ratio 1:1 of diffused mixture. Which one are not correct if diffusion is made at same P and T in each operation? (I) Eight operation are needed to get 1:1 molar ratio. (II) Rate of diffusion for SO_(2):O_(3) after eight operations in 0.707. (III) Six operations are needed to get 2:1 molar ratio for SO_(2) and O_(2) in diffusion mixture. (IV) Rate of diffusion for SO_(2) and O_(2) after six operations is 2.41.
Answer» `I`, `II`, `III` `II`, `III` `I`, `III` `IV` Solution :`(f_(1))^(X)=(n'_(SO_(2)))/(n'_(O_(2)))xx(n_(O_(2)))/(n_(SO_(2)))`, where `n_(SO_(2))` and `n_(O_(2))` are moles present initially. or `XlogF_(1)=log[(n'_(SO_(2)))/(n'_(O_(2)))xx(n_(O_(2)))/(n_(SO_(2)))]` `:. Xlogsqrt((M_(O_(2)))/(M_(SO_(2))))=log[(n'_(SO_(2)))/(n'_(O_(2)))xx(n_(O_(2)))/(n_(SO_(2)))]` `X logsqrt((32)/(64))=log(1)/(1)xx(1)/(16)` `:. X=8`, also `(n_(1))/(n_(2))=(r_(1))/(r_(2))=sqrt((32)/(64))=0.707` If `X=6`, then `6 log sqrt((32))/(64))=log[(n'_(SO_(2)))/(n'_(O_(2)))xx(n_(O_(2)))/(n_(SO_(2)))]` `=log[(n'_(SO_(2)))/(n'_(O_(2)))xx(1)/(16)]` `(n'_(SO_(2)))/(n'_(O_(2)))=2:1` RATE of diffusion is `(r_(1))/(r_(2))=sqrt((M_(2))/(M_(1)))`, i.e., `0.707` in each OPERATION.