If 0.15 g of a solute dissolved in 15 g of solvent is boiled at a temperature higher by 0.216° C, than that of the pure solvent, the molecular weight of the substance is: (Molal elevation constant for the solvent is 216° C )

Question

If 0.15 g of a solute dissolved in 15 g of solvent is boiled at a temperature higher by 0.216° C, than that of the pure solvent, the molecular weight of the substance is: (Molal elevation constant for the solvent is 216° C ) (A) 10 (B) 10.1 (C) 100 (D) 1.01

Tardigrade Admin · Accepted Answer

Key Idea: Use the following formula to find molecular formula of solute.

M = \frac{K _{b} \times w \times 1000}{Δ T _{b} \times W}

where

M =

molecular mass of solute

w =

weight of solute

= 0.15 g

W =

weight of solvent

= 15 g

Δ T_{b} =

elevation in boiling point

= 0.21 6^{\circ} CM

= \frac{2.16 \times 0.15 \times 100}{0.216 \times 15} = 100

Q. If 0.15g of a solute dissolved in 15g of solvent is boiled at a temperature higher by 0.216∘C, than that of the pure solvent, the molecular weight of the substance is: (Molal elevation constant for the solvent is 216∘C )

10

10.1

100

1.01

Key Idea: Use the following formula to find molecular formula of solute. M=ΔTb​×WKb​×w×1000​ where M= molecular mass of solute w= weight of solute =0.15g W= weight of solvent =15g ΔTb​= elevation in boiling point =0.216∘CM =0.216×152.16×0.15×100​=100

Q. If $0.15 g$ of a solute dissolved in $15 g$ of solvent is boiled at a temperature higher by $0.21 6^{\circ} C$ , than that of the pure solvent, the molecular weight of the substance is: (Molal elevation constant for the solvent is $21 6^{\circ} C$ )