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A solute A dimerizes in water. The boiling point of a 2 molal solution of A is 100.52^{\circ}C. The percentage associated of A is ______ (Rounded off to the nearest integer) [ Use: K_b for water =0.52 K kg mol^{-1}  Boiling point of water = 100 ^{\circ}C]
 

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So,

i = \frac{\text { Initial moles } }{\text { Final moles }}

i=\frac{1-\alpha+1-\alpha+\alpha}{1+1}

i=1-\frac{\alpha}{2}

Now,

\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{T}_{\mathrm{b}}-\mathrm{T}_{\mathrm{b}}^{0}

\Delta \mathrm{T}_{\mathrm{b}}=100.52 -100

\Delta \mathrm{T}_{\mathrm{b}}=0.52^{\circ} \mathrm{C}

\because \Delta \mathrm{T}_{\mathrm{b}}=\mathrm{i} \mathrm{K}_{\mathrm{b}} \times \mathrm{m}

0.52=\left(1-\frac{\alpha}{2}\right) \times 0.52 \times 2

So, \alpha=1

So, percentage association = 100%.

Ans = 100

Posted by

Kuldeep Maurya

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