Arrange the following solutions in order of their increasing boiling points

$\left(\mathrm{i}\right){10}^{-4}\mathrm{M}\mathrm{NaCl}\\ \left(\mathrm{ii}\right){10}^{-4}\mathrm{M}\mathrm{Urea}\\ \left(\mathrm{iii}\right){10}^{-3}\mathrm{M}\mathrm{NaCl}\\ \left(\mathrm{iv}\right){10}^{-2}\mathrm{M}\mathrm{NaCl}$

Question

Arrange the following solutions in order of their increasing boiling points

$\left(\mathrm{i}\right){10}^{-4}\mathrm{M}\mathrm{NaCl}\\ \left(\mathrm{ii}\right){10}^{-4}\mathrm{M}\mathrm{Urea}\\ \left(\mathrm{iii}\right){10}^{-3}\mathrm{M}\mathrm{NaCl}\\ \left(\mathrm{iv}\right){10}^{-2}\mathrm{M}\mathrm{NaCl}$

YESPYQ · Accepted Answer

Correct answer: c) (ii) < (i) < (iii) < (iv). $∆{\mathrm{T}}_{\mathrm{b}}={\mathrm{iK}}_{\mathrm{b}}\mathrm{m}$ $∆{\mathrm{T}}_{\mathrm{b}}\alpha \mathrm{i}\times \mathrm{m}$ For NaCl " i " = 2 and Urea " i " = 1 Higher the product of vant Hoff factor and concentration higher the boiling point. i × C = (i) 2 × 10 –4 (ii) 1 × 10 –4 (iii) 2 × 10 –3 (iv) 2 × 10 –2 Hence the order for boiling point is: (ii) < (i) < (iii) < (iv)

Arrange the following solutions in order of their increasing boiling points \(\left(\mathrm{i}\right){10}^{-4}\mathrm{M}…

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