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 P4 is the molecular formula of P
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 P4 is the molecular formula of P