Q.
0 .15 g of a substance dissolved in 15 g of a solvent higher by 0.216∘C than that of the pure solvent. Find out the molecular weight of the substance. (Kbforsolventis2.16∘C)
Elevation in boiling point, ΔT=KbXmb
Here, Kb=2.16,ΔT=.216 mb=ΔT/Kb=>.216/2.16=0.1
Now, molality, mb= Moles of solute/Mass of solvent(in kg) 0.1= Moles of solute/ 0.015 ⇒ Moles of solute =0.0015
Finally, Moles= Mass (in gm)/ Mol. Wt 0.0015=0.15 / Mol wt
Mol w t =100