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COMEDK · Chemistry · 17. Solutions

If the depression in freezing point of an aqueous solution containing a solute, which is neither dissociated nor associated, is \(\mathrm{a} \mathrm{K}\) with \(\mathrm{K}_{\mathrm{f}}=\mathrm{b ~ K ~ k g ~ m o l}{ }^{-1}\), what would be the elevation in boiling point \((\mathrm{X})\) for this solution if its \(\mathrm{K}_{\mathrm{b}}=\mathrm{\mathrm { ~K } \mathrm { ~K } \mathrm { ~kg } \mathrm { ~mol }}{ }^{-1}\) ?

  1. A \(\mathrm{X}=\mathrm{c} \times \dfrac{b}{a}\)
  2. B \(X=c \times \dfrac{a}{b}\)
  3. C \(\mathrm{X}=2 \mathrm{c} \times \dfrac{b}{a}\)
  4. D \(X=c \times \dfrac{a}{2 b}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(X=c \times \dfrac{a}{b}\)

Step-by-step Solution

Detailed explanation

The depression in freezing point is given by the formula \(\Delta T_f = K_f \times m\), where \(m\) is the molality of the solution. Given \(\Delta T_f = a\) and \(K_f = b\), we have \(a = b \times m\), which implies \(m = \dfrac{a}{b}\). The elevation in boiling point is given…
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