If 0.2g of a gas 'X' occupies a volume of 440 ml and if 0.1 g of `CO_(2)` gas occupies a volume of 320 ml at the same temperature and pressure , X could be

Options

A. `O_(2)`
B. NO
C. `C_(4) H_(10)`
D. `SO_(2)`

Answer

D

Explanation

At same T and p,
`( V_(1))/( V_(2)) = ( n_(1))/(n_(2)) = ( W_(1)//M_(1))/(W_(2)//M_(2))`
`( 440 ml )/( 320ml ) = ( 0.2 //M_(1))/( 0.1 // 44 )`
`:. M _(1) = 66g mol^(-1)`
Hence gas X could be `SO_(2)`