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Elevation in Boiling Point

The elevation...

Question

The elevation in boiling point of a solution of $13.44 g$ of ${C u C l}_{2}$ , (molecular weight $= 134.4$ , $K_{b} =$ 0.52 K molarity ) in 1 kg water using the following information will be:

0.16

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0.05

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0.1

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0.2

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Solution

The correct option is A $0.16$
$Δ T_{b} = i \times M o l a l i t y \times K_{b} ⟹ Δ T_{b} = 3 \times \frac{\frac{13.44 g m}{134.4 g / m o l}}{1 k g} \times 0.52 = 3 \times \frac{1}{10} \times 0.52$

$Δ T_{b} = 3 \times 0.052 = 0.156 \approx 0.16 K$

Elevation in boiling point $= 0.16 K$

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Similar questions

Q. The elevation in boiling point of a solution of 13.44 g of

C u C l_{2}

in 1 kg of water using the following information will be (molecular weight of

C u C l_{2}

is 134.4 and

K_{b} = 0.52 K m^{- 1}

)

Q. The elevation in boiling point of a solution of

13.44 g

C u C l_{2}

in 1 kg of water using the following information will be ( Molecular weight of

C u C l_{2} = 134.4

and

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

The elevation in boiling point of a solution of 13.44g of CuCl₂ in 1kg of water using the following information will be (Molecular weight of CuCl₂= 134.4 and K_b= 0.52K molal^-1)

Q. The elevation in boiling point of a solution of 13.44 g of

C u C l_{2}

(Molar mass 134.4) in 1 kg water is:

[K_{b} = 0.52

K molality

^{- 1}]

Q. The elevation in boiling point, when 13.44 g of freshly prepared

C u C l_{2}

are added to one kilogram of water is:

[Some useful data,

K_{b}

= 0.52

k g K m o l^{- 1}

. Molecular weight of

C u C l_{2} =

134.4 gm]

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The elevation in boiling point of a solution of 13.44g of CuCl2, (molecular weight =134.4, Kb= 0.52 K molarity ) in 1 kg water using the following information will be:

The correct option is A 0.16ΔTb=i×Molality×Kb⟹ΔTb=3×13.44gm134.4g/mol1kg×0.52=3×110×0.52ΔTb=3×0.052=0.156≈0.16 KElevation in boiling point =0.16 K

The elevation in boiling point of a solution of $13.44 g$ of ${C u C l}_{2}$ , (molecular weight $= 134.4$ , $K_{b} =$ 0.52 K molarity ) in 1 kg water using the following information will be:

The correct option is A $0.16$
$Δ T_{b} = i \times M o l a l i t y \times K_{b} ⟹ Δ T_{b} = 3 \times \frac{\frac{13.44 g m}{134.4 g / m o l}}{1 k g} \times 0.52 = 3 \times \frac{1}{10} \times 0.52$

$Δ T_{b} = 3 \times 0.052 = 0.156 \approx 0.16 K$

Elevation in boiling point $= 0.16 K$