JEE Mains · Chemistry · STD 12 -1. Solution and colligative properties
\(AB _{2}\) is \(10 \%\) dissociated in water to \(A ^{2+}\) and \(B ^{-}\). The boiling point of a \(10.0\) molal aqueous solution of \(AB _{2}\) is ............ \({ }^{\circ} C\). (Round off to the Nearest Integer). [Given : Molal elevation constant of water \(K _{ b }=0.5\, \,K\, kg\, mol ^{-1}\) boiling point of pure water \(\left.=100^{\circ} C \right]\)
- A \(201\)
- B \(105\)
- C \(102\)
- D \(106\)
Answer & Solution
Correct Answer
(D) \(106\)
Step-by-step Solution
Detailed explanation
\(AB _{2} \rightarrow A ^{2+}+2 B ^{-}\) \(\begin{array}{llll} t =0 a 0 0\end{array}\) \(\begin{array}{llll} t = t a - a \alpha a \alpha 2 a \alpha\end{array}\) \(n _{ T }= a - a \alpha+ a \alpha+2 a \alpha\) \(= a (1+2 \alpha)\) so \(i =1+2 \alpha\) Now…
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