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Molal elevation constant values of following alcohols are in the order:$CH_3CH_2CH_2CH_2OH > (CH_3)_2CH-CH_2OH > (CH_3)_3C-OH$ . Explain in brief.

1 Answer

Moving from $CH_3CH_2CH_2CH_2OH$ to $(CH)_3C-OH$ via $(CH_3)_2CHCH_2OH,$ , the branching increases, the surface area decreases resulting into decrease in Vander Waals' force and hence boiling point.
$K_b=\large\frac{M_{A}RT_b^{0^2}}{\Delta_{vap}H\times 1000}$
Molecular weight (M) remains the same for all the three isomeric alcohols.
$\large\frac{\Delta _{vap}H}{T_b}$ is almost constant for all the liquids of almost similar kind of association.
Therefore, $K_b \propto M_A$
answered Jun 11, 2014 by sreemathi.v
 

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