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

Molality

An aqueous so...

Question

An aqueous solution of $N a C l$ freezes at $- {0.816}^{\circ} C$ . The elevation in boiling point of this solution is:

Given that $K_{b (H_{2} O)} = 0.512$ K kg mol $^{- 1}$ and $K_{f (H_{2} O)} = 1.86$ K kg mol $^{- 1}$ .

0.0585^{0} C

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

0.2246^{0} C

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

1.864^{0} C

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

0.0265^{0} C

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

Open in App

Solution

The correct option is C $0.2246^{0} C$
$Δ T_{f} = i K_{f} m$ ....(1)

$Δ T_{b} = i K_{b} m$ .....(2)

Divide equation (2) by equation (1)

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

$\frac{Δ T_{b}}{0.816} = \frac{0.512}{1.86}$

$Δ T_{b} = {0.2246}^{0} C$

Hence, the correct option is $B$ .

Suggest Corrections

Similar questions

- {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})

{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}

- {0.186}^{o}

K_{f}

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

K_{b}

0.512 K k g m o l^{- 1}

K_{f} = 1.86

^{- 1}

K_{b} = 0.512

^{- 1}

298 K = 23.756

- {0.186}^{0} C

K_{f} = {1.86}^{0} m o l^{- 1} k g

K_{b} = {0.512}^{0} m o l^{- 1} k g

Join BYJU'S Learning Program