0.1g of carbon dioxide occupies a volume of 320cc at certain conditions. Under similar conditions 0.2g of 3 dioxide of element 'X' occupies 440cc. Calculate the atomic weight of 'X'.

Explanation

Applying Avogadro.s law,-`(n_1)/(n_2) = (V_1)/(V_2)`
But number of moles = n = w/M
`n=(" weight ")/("gram molecular weight ") " " (V_1)/(V_2) = (w_1 M_2)/(w_2 M_1)`
Molecular weight of dioxide of X =
`M_2 (V_1)/(V_2) xx (w_2 M_1) /(w_1) = ( 320)/( 440 ) xx (0.2 )/( 0.1 ) xx 44 =64`
Molecular weight of `XO_2 = 64`
Therefore, atomic weight of X = 32