An aqueous solution boils at 100.2° C . At which temperature this will freeze. (Kb=0.5° C/m,Kf=1.9° C/m)

Question

An aqueous solution boils at 100.2° C . At which temperature this will freeze. (Kb=0.5° C/m,Kf=1.9° C/m)  (A)  +0.76  (B)  -0.76  (C)  -0.38  (D)  +0.38

Tardigrade Admin · Accepted Answer

∵ Tb=100.2° C ∴ Δ Tb=100.2-100=0.2° C Δ Tb=Kb× m m=(Δ Tb/Kb)=(0.2/0.5)=0.4 Δ Tf=Kf× m=1.9× 0.4 =0.76° C Δ Tf=To-Δ Tf=0-0.76 =-0.76° C

Q. An aqueous solution boils at $100.2^{\circ} C$ . At which temperature this will freeze. $(K_{b} = 0.5^{\circ} C / m, K_{f} = 1.9^{\circ} C / m)$

$+ 0.76$

$- 0.76$

$- 0.38$

$+ 0.38$

$∵$ $T_{b} = 100.2^{\circ} C$
$∴$ $Δ T_{b} = 100.2 - 100 = 0.2^{\circ} C$
$Δ T_{b} = K_{b} \times m$
$m = \frac{Δ T _{b}}{K _{b}} = \frac{0.2}{0.5} = 0.4$
$Δ T_{f} = K_{f} \times m = 1.9 \times 0.4$
$= 0.76^{\circ} C$
$Δ T_{f} = T_{o} - Δ T_{f} = 0 - 0.76$
$= - 0.76^{\circ} C$

Q. An aqueous solution boils at 100.2∘C . At which temperature this will freeze. (Kb​=0.5∘C/m,Kf​=1.9∘C/m)

+0.76

−0.76

−0.38

+0.38