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. (Kb for solvent is 2.16 ° C)

Question

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. (Kb for solvent is 2.16 ° C) (A) 1.01 (B) 10.1 (C) 100 (D) 10

Tardigrade Admin · Accepted Answer

Elevation in boiling point, Δ T = K b Xm b Here, K b =2.16, Δ T =.216 m b =Δ T / K b =>.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

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. (Kb​forsolventis2.16∘C)

1.01

10.1

100

10

Elevation in boiling point, ΔT=Kb​Xmb​ 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

Q. 0 .15 g of a substance dissolved in 15 g of a solvent higher by 0.216 $^{\circ}$ C than that of the pure solvent. Find out the molecular weight of the substance. $(K_{b} f or so l v e n t i s 2.1 6^{\circ} C)$