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

Elevation in Boiling Point

An aqueous so...

Question

An aqueous solution of urea freezes at $- {0.186}^{o}$ C. $K_{f}$ for water = $1.86 K k g m o l^{- 1}$ , $K_{b}$ for water = $0.512 K k g m o l^{- 1}$ . The boiling point of urea solution will be :

373.065

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

373.186

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

373.512

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

373.0512

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

Open in App

Solution

The correct option is B $373.0512$ K
Given that,

$K_{f} = 1.86 K k g / m o l; K_{b} = 0.512 K k g / m o l; T_{f, s o l u t i o n} = - {0.186}^{o} C$ and $T_{f, w a t e r} = 0^{o} C$

$Δ T_{f} = 0.186 K$

Since,

$\frac{Δ T_{f}}{K_{f}} = \frac{Δ T_{b}}{K_{b}} = m$

$Δ T_{b} = \frac{0.186}{1.86} \times 0.512 = 0.0512 K$

As, $T_{b, w a t e r} = 373 K,$ and

$Δ T_{b} = 0.0512 K$
Since there is an elevation in boiling point upon addition of solute

$∴ T_{b, s o l u t i o n} = 373 + 0.0512 = 373.0512 K$

Hence option D is correct.

Suggest Corrections

Similar questions

K_{f} = 1.86 K k g m o l^{- 1}, K_{b} = 0.512 K k g m o l^{- 1}

- {0.186}^{o} C

(K_{f} = 1.86 K k g m o l^{- 1}, K_{b} = 0.512 K k g m o l^{- 1})

{100.15}^{o} C .

K_{f}

K_{b}

m o l^{- 1}

m o l^{- 1}

K_{f} = 1.86

^{- 1}

K_{b} = 0.512

^{- 1}

298 K = 23.756

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

Join BYJU'S Learning Program