0.3 molal aqueous solution of $C u C l_{2}$ is $75 %$ ionized. If $K_{b}$ for water is $0.52 K k g m o l^{- 1}$ , the elevation in boiling point of solution is:

0.39 K

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{0.39}^{\circ} C

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Both (A) and (B)

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None of these

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Solution

The correct option is C Both (A) and (B)
$Δ T_{b} = i \times K_{b} \times m$
$i = 1 + (n - 1) α$
for $C u C l_{2} n = 3, α = \frac{3}{4}$
$= 1 + (2) \times \frac{3}{4} = \frac{5}{2}$
$Δ T_{b} = \frac{5}{2} \times 0.52 \times 0.3 = 0.39 K o r {0.39}^{\circ} C$

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Similar questions

A B

75 %

1.82 K

= 0.52 {K kg mol}^{- 1}

C u C l_{2}

[K_{b} = 0.52

^{- 1}]

A_{2} B_{3}

60 %

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

If 2 moles of sugar are dissolved in 1000g of water, the elevation in boiling point of the resulting solution is: (K_b = 0.52 K kg mol^-1):

13.44 g

C u C l_{2}

C u C l_{2} = 134.4

K_{b} = 0.52 K m o l a l^{- 1}

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