NEET UG 2005 — Chemistry Physical Chemistry

$$\begin{aligned} \Delta \mathrm{T}_f & =\mathrm{K}_f m \ldots(1) \\ \Delta \mathrm{T}_b & =\mathrm{K}_b m \ldots(2) \\ \Rightarrow \quad \frac{\Delta \mathrm{T}_f}{\Delta \mathrm{T}_b} & =\frac{\mathrm{K}_f}{\mathrm{~K}_b} \ldots(3) \end{aligned}$$ΔTf​ΔTb​⇒ΔTb​ΔTf​​​=Kf​m…(1)=Kb​m…(2)= Kb​Kf​​…(3)​ $$\begin{aligned} & \text { b.p. of water }=100^{\circ} \mathrm{C} ; \mathrm{K}_f=1.86 \mathrm{~kg} \\ & \mathrm{~mol}^{-1} \\ & \text { b.p. of urea in water }=100.18^{\circ} \mathrm{C} ; \mathrm{K}_b \\ & =0.512 \mathrm{~kg} \mathrm{~mol}^{-} \\ & \Rightarrow \quad \Delta \mathrm{T}_b=0.18 \\ & \text { f.p. of water }=0^{\circ} \mathrm{C} \\ & f . p . \text { of urea in water }=-\mathrm{T}^{\circ} \mathrm{C} \\ & \Rightarrow \quad \Delta \mathrm{T}_f=\mathrm{T} \\ & \Rightarrow \text { from eq. }(3) \\ & \qquad \frac{\mathrm{T}}{0.18}=\frac{1.86}{0.512} \\ & \Rightarrow \quad \mathrm{T}=0.6539 \\ & \Rightarrow f . p . \text { of urea in water }=-0.654^{\circ} \mathrm{C} \end{aligned}$$​ b.p. of water =100∘C;Kf​=1.86 kg mol−1 b.p. of urea in water =100.18∘C;Kb​=0.512 kg mol−⇒ΔTb​=0.18 f.p. of water =0∘Cf.p. of urea in water =−T∘C⇒ΔTf​=T⇒ from eq. (3)0.18T​=0.5121.86​⇒T=0.6539⇒f.p. of urea in water =−0.654∘C​