A solution containing `28`g of phosphorus in `315`g `CS_(2)(b.p.46.3^(@)C`boils at `47.98^(@)C`. If `k_(b)`for `CS_(2)`is `2.34`K kg `mol^(-1)`. The formula of phosphorus is (at .massof P=`31`).

Options

A. `P_(6)`
B. `P_(4)`
C. `P_(3)`
D. `P_(2)`

Answer

B

Explanation

`DeltaT_(b)=mK_(b)=(w)/(M)xx(1000)/(W)xxK_(b)`
`DeltaT_(b)=47.98-46.3=1.68`
`1.68=(28)/(M)xx(1000)/(315)xx2.38`
`M=(28xx1000xx2.38)/(315xx1.68)=125.92`
atomicity `=("Mol.wt.")/("At.wt.")=(125.92)/(31)=4.02`
So. Molecule is `=P_(4)`.