If any solute - A- dimerises in water at 1 atm pressure and the boiling point of this solution is 100-52-C- If 2 moles of A is added to 1 kg of water and kb for water is 0-52-C-molal- calculate the percentage association of A-

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Elevation in Boiling Point

If any solute...

Question

If any solute ' $A$ ' dimerises in water at 1 atm pressure and the boiling point of this solution is ${100.52}^{\circ} C$ . If 2 moles of $A$ is added to 1 kg of water and $k_{b}$ for water is ${0.52}^{\circ} C / m o l a l$ , calculate the percentage association of $A$ ?

50%

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30%

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25%

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100%

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Solution

The correct option is A 50%
$2 A ⇌ A_{2}$
Elevation of boiling point =100.52-100=0.52
$Δ T_{b} = i \cdot K_{b} \cdot (m o l a l i t y)$
Molality = $\frac{2 m o l e s}{1 k g} = 2 m o l a l$
$0.52 = i (0.52) (2)$
$i = 0.5$
$∴$ Degree of association $= 0.5 \times 100 = 50$ %

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Similar questions

Q. A solute A dimerizes in water. The boiling point of a 2 molal solution of A is 100.52 ºC. The percentage association of A is________.(Round off to the nearest integer.)

[Use :

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for water =

0.52 K k g m o l^{- 1}

Boiling point of water = 100 ºC]

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If any solute 'A' dimerises in water at 1 atm pressure and the boiling point of this solution is 100.52∘C. If 2 moles of A is added to 1 kg of water and kb for water is 0.52∘C/molal, calculate the percentage association of A?

The correct option is A 50%2A⇌A2Elevation of boiling point =100.52-100=0.52ΔTb=i⋅Kb⋅(molality)Molality = 2moles1kg=2molal0.52=i(0.52)(2)i=0.5∴ Degree of association =0.5×100=50%

If any solute ' $A$ ' dimerises in water at 1 atm pressure and the boiling point of this solution is ${100.52}^{\circ} C$ . If 2 moles of $A$ is added to 1 kg of water and $k_{b}$ for water is ${0.52}^{\circ} C / m o l a l$ , calculate the percentage association of $A$ ?

The correct option is A 50%
$2 A ⇌ A_{2}$
Elevation of boiling point =100.52-100=0.52
$Δ T_{b} = i \cdot K_{b} \cdot (m o l a l i t y)$
Molality = $\frac{2 m o l e s}{1 k g} = 2 m o l a l$
$0.52 = i (0.52) (2)$
$i = 0.5$
$∴$ Degree of association $= 0.5 \times 100 = 50$ %