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Elevation in Boiling Point

1.0 molal aqu...

Question

1.0 molal aqueous solution of an electrolyte $A_{2} B_{3}$ is $60 %$ ionized. The boiling point of the solution at 1 atm is:
$[K_{b (H_{2} O)} = 0.52 K k g m o l^{- 1}]$

274.76

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377

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376.4

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374.72

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Solution

The correct option is D $374.72$ K
The vant Hoff factor (i) is

$i = \frac{n (o b s r v e d)}{n (t h e o r e t i c a l)} = \frac{(2 + 3) \times 0.6}{1} = 3$

The expression for the elevation in boiling point is

$Δ T_{b} = k_{b} \times m \times i = 0.52 \times 1.0 \times 3 = 1.56 K$

The boiling point of 1.0 molal aqueous solution is $373.15 K + 1.56 = 374.72 K$

Hence, the correct option is $D$

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Similar questions

1 m o l a l

A_{2} B_{3}

60 %

1 a t m

K

K_{b}

(H_{2} O) = 0.52 K k g m o l^{- 1}]

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H_{2} O = 0.52 K k g / m o l]

0.1 molal

A B_{4}

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K_{f} = 1.86 K m o l^{- 1} k g

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