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\(1 \mathrm{~mol}\) each of the following compounds is dissolved in \(1 \mathrm{~L}\) of solution. Which will have the largest \(\Delta T_b\) value?
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The correct answer is:
\(\mathrm{HI}\)
Bond dissociation energy increases in the order.
\(\mathrm{HI} < \mathrm{HBr} < \mathrm{HCl} < \mathrm{HF}\)
\(\mathrm{HF}\) is most stable halogen acid and \(\mathrm{HI}\) is the least. Higher the bond dissociation energy, lower is the degree of ionisation.
As we know \(\Delta T_b=i K_b m\).
Hence \(\Delta T_b\) value is largest in HI. All colligative properties depend on number of particles.
\(\mathrm{HI} < \mathrm{HBr} < \mathrm{HCl} < \mathrm{HF}\)
\(\mathrm{HF}\) is most stable halogen acid and \(\mathrm{HI}\) is the least. Higher the bond dissociation energy, lower is the degree of ionisation.
As we know \(\Delta T_b=i K_b m\).
Hence \(\Delta T_b\) value is largest in HI. All colligative properties depend on number of particles.
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