You visited us 1 times! Enjoying our articles? Unlock Full Access!

Relative Lowering of Vapour Pressure

The elevation...

Question

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:
$[K_{b} = 0.52 K {molality}^{- 1}]$

0.16

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0.05

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $0.16$
The elevation in boiling point of a solution is 0.16

$Δ T_{b} = i K_{f} m$
$Δ T_{b} = \frac{1000 \times 0.52 \times 13.44}{134.4 \times 1000} (1 + 2 α)$
$∴ Δ T_{b} = 0.156 \approx 0.16$ $(∵ α = 1)$

Hence, the correct option is $A$

Suggest Corrections

Similar questions

C u C l_{2}

[K_{b} = 0.52

^{- 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}

C u C l_{2}

C u C l_{2}

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

13.44 g

C u C l_{2}

1 k g

C u C l_{2}

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

C u C l_{2} = 134.4 g / m o l

13.44 g

{C u C l}_{2}

= 134.4

K_{b} =

Join BYJU'S Learning Program