Question

$44.8\, mL$ of a vapour under $STP$ ( $P = 1$ atmosphere, $T = 273\, K$ ) conditions has a mass of $320\, mg$ . This could be the vapour of (assume ideal behaviour for the vapour) (Atomic masses of $H$ , $Br$ , $C$ and $I$ are respecti- vely $1, 80, 12$ and $127 \,g \, mol^{-1}$ )

• A. benzene
• B. toluene
• C. bromine
• D. iodine

Answer 1

Answer: (C)

bromine

Answer 2

$44.8\,mL$ has a mass of $320\,mg$
$\therefore 22400\,mL$ will have a mass of $\frac{320}{44.8}\times 22400$
$=160000\,mg=160\,g$
Hence, the vapours are of bromine $(Br_{2})$