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

Elevation in Boiling Point

For an aqueou...

Question

For an aqueous solution, freezing point is $- {0.186}^{0} C$ . Elevation of the boiling point is the same solution is : ( $K_{f} = {1.86}^{0} m o l^{- 1} k g$ and $K_{b} = {0.512}^{0} m o l^{- 1} k g$ )

{0.186}^{0} C

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

{0.0512}^{0} C

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

{1.86}^{0} C

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

{5.12}^{0} C

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

Open in App

Solution

The correct option is B ${0.0512}^{0} C$
The depression in freezing point is given as follow:
$Δ T_{f} = - K_{f} m$
$m = \frac{Δ T_{f}}{- K_{f}} = \frac{- 0 \cdot 186}{1 \cdot 86}$
$m = 0 \cdot 1$
Now, elevation of the boiling point can be calculated as follow:
$Δ T_{b} = K_{b} m$
$= 0 \cdot 512 \times 0.1$
$= 0 \cdot 0512^{0} C$

Suggest Corrections

Similar questions

^{o}

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

K_{b} = {0.512}^{o} C m o l^{- 1} k g

{0.186}^{\circ} C

K_{f}

m o l e^{- 1}

K_{b}

m o l e^{- 1}

- {0.156}^{o} C

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

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

K_{f}

{1.86}^{0} m o l^{- 1} K g

Δ T_{f}

{0.186}^{o} C

K_{f}

= {1.86}^{o} C k g m o l^{- 1}

K_{b}

= {0.512}^{o} C k g m o l^{- 1}

Join BYJU'S Learning Program