The freezing point of a 0.01 m aqueous glucose solution at 1 atmosphere is -0.18° C . To it, an addition of equal volume of 0.002 m glucose solution will produce a solution with freezing point of nearly

Question

The freezing point of a 0.01 m aqueous glucose solution at 1 atmosphere is -0.18° C . To it, an addition of equal volume of 0.002 m glucose solution will produce a solution with freezing point of nearly (A)  -0.036° C  (B)  -0.108° C  (C)  -0.216° C  (D)  -0.422° C

Tardigrade Admin · Accepted Answer

: Depression in freezing point of 0.01 M aqueous glucose solution =0.18° C Here, Δ Tf=Kf× m Kf=(0.18/0.01)=18 Molality of solution obtained by mixing equal volumes of 0.01 m and 0.002 m solution =(0.01+0.002/2)=(0.012/2)=0.006m For 0.006 m glucose solution, Δ Tf=18× 0.006=0.108° C .

Q. The freezing point of a 0.01 m aqueous glucose solution at 1 atmosphere is $- 0.18^{\circ} C$ . To it, an addition of equal volume of 0.002 m glucose solution will produce a solution with freezing point of nearly

$- 0.036^{\circ} C$

$- 0.108^{\circ} C$

$- 0.216^{\circ} C$

$- 0.422^{\circ} C$

Q. The freezing point of a 0.01 m aqueous glucose solution at 1 atmosphere is −0.18∘C . To it, an addition of equal volume of 0.002 m glucose solution will produce a solution with freezing point of nearly

−0.036∘C

−0.108∘C

−0.216∘C

−0.422∘C

: Depression in freezing point of 0.01 M aqueous glucose solution =0.18∘C Here, ΔTf​=Kf​×m Kf​=0.010.18​=18 Molality of solution obtained by mixing equal volumes of 0.01 m and 0.002 m solution =20.01+0.002​=20.012​=0.006m For 0.006 m glucose solution, ΔTf​=18×0.006=0.108∘C .

Q. The freezing point of a 0.01 m aqueous glucose solution at 1 atmosphere is $- 0.18^{\circ} C$ . To it, an addition of equal volume of 0.002 m glucose solution will produce a solution with freezing point of nearly

$- 0.036^{\circ} C$

$- 0.108^{\circ} C$

$- 0.216^{\circ} C$

$- 0.422^{\circ} C$