a solution of 28g phosphorous in 315g CS2(boiling point=46.3deg C) boils at 47.9deg C(kb for CS2 is 2.34). What will be the molecular formula of phosphorous?

ΔTb=kbm
​where m is molality
m=no. of molesof solute/kg of solvent
no. of moles =w/M =28 / M
m =(28/M)/0.315  moles/kg

ΔTb=kbm
​1.6=2.34 *[​(28/M)/0.315 ]

M = 130 i.e molar mass of phosphorous
Thus no. of atoms of P present =130/31 =4.19 =approx 4

Thus Pis the molecular formula of P
 

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