If 0.15 g of a solute dissolved in 15 got solvent is boiled at a temperature higher by 0.216°C , than that of the pure solvent, the molecular weight of the substance, (molal elevation constant for the solvent is 2.16° C ):

Question

If 0.15 g of a solute dissolved in 15 got solvent is boiled at a temperature higher by 0.216°C , than that of the pure solvent, the molecular weight of the substance, (molal elevation constant for the solvent is 2.16° C ): (A) 100 (B) 10.1 (C) 10 (D) 1.01

Tardigrade Admin · Accepted Answer

Molecular weight of solute is calculated by using following formula.

M_{B} = \frac{K _{b} \times W _{B} \times 1000}{Δ T _{b} \times W _{a}}

where,

M_{B} =

molecular v/eight of solute

W_{B} =

weight of solute

= 0.15 g

Δ T_{b} =

elevation in boiling point

= 0.21 6^{\circ} C W_{a} =

weight of solvent

= 15.0 g

K_{b} = 2.16

∴ M_{B} = \frac{2.16 \times 0.15 \times 1000}{0.216 \times 15}

= 100

Q. If 0.15g of a solute dissolved in 15 got solvent is boiled at a temperature higher by 0.216∘C , than that of the pure solvent, the molecular weight of the substance, (molal elevation constant for the solvent is 2.16∘C ):

100

10.1

10

1.01

Q. If $0.15 g$ of a solute dissolved in $15$ got 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, (molal elevation constant for the solvent is $2.1 6^{\circ} C$ ):