A solution of urea (mol. mass .56 ~g ~mol^ -1 ) boils at 100.18^ C at the atmospheric pressure. If K_f and K_b for water for 1.86 and 0.512 Kkmol^ -1 respectively, the above solution will freeze at:

Correct answer: B. -0.654^ C aligned Δ T_ f & =K_ f m (1) Δ T_ b & =K_ b m (2) Δ T_f Δ T_b & = K_f K_b (3) aligned aligned & b.p. of water =100^ ^ C; K_ f=1.86 kg & mol^ -1 & b.p. of urea in water =100.18^ ^ C; K_ b & =0.512 kg mol^ - & Δ T_ b=0.18 & f.p. of water =0^ ^ C & f . p . of urea in water =-T^ ^ C & Δ T_ f=T & from eq. (3) & T 0.18 =(1.86)/(0.512) & T=0.6539 & f . p . of urea in water =-0.654^ ^ C aligned

NEET 2005

Exam: NEET
Year: 2005
Subject: Chemistry
Chapter: Solutions

NEET 2005 · Chemistry · Solutions

A solution of urea (mol. mass .56 ~g ~mol^ -1 ) boils at 100.18^ C...

A solution of urea (mol. mass

\left.56 \mathrm{~g} \mathrm{~mol}^{-1}\right)

boils at

100.18^{\circ} \mathrm{C}

at the atmospheric pressure. If

\mathrm{K}_f and \mathrm{K}_b

for water for 1.86 and

0.512 \mathrm{Kkmol}^{-1}

respectively, the above solution will freeze at:

Options

A $0.654^{\circ} \mathrm{C}$
B $-0.654^{\circ} \mathrm{C}$
C $6.54^{\circ} \mathrm{C}$
D $-6.54^{\circ} \mathrm{C}$

Correct answer

B. $-0.654^{\circ} \mathrm{C}$

Step-by-step solution

\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}

\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}

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