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

Depression in Freezing Point

An aqueous so...

Question

An aqueous solution of a non-volatile solute boils at ${100.17}^{o} C .$ At what temperature will the solution freeze?
(Given: $K_{b} = 0.512 K k g m o l^{- 1}$ and $K_{f} = 1.86 K k g m o l^{- 1}$ )

Open in App

Solution

We know that,
$△ T_{b} = m o l a l i t y \times K_{b}$
$0.17 = m o l a l i t y \times 0.512$
Molality of the solution $= \frac{0.17}{0.512} m$
Let depression in freezing point be $△ T_{f}$
$△ T_{f} = m o l a l i t y \times K_{f}$
$= \frac{0.17}{0.512} \times 1.86 = {0.62}^{o} C$
Thus, the freezing point of the solution
$= 0.00 - 0.62 = - {0.62}^{o} C$

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}

K_{f}

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

K_{b}

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

- {0.93}^{0} C

k_{f} = 1.86 K k g m o l^{- 1}, k_{b} = 0.52 K k g m o l^{- 1}

{100.17}^{\circ} C .

^{\circ} C

[K_{b}

K_{f}

0.512

{1.86}^{\circ} C / m

]

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

Join BYJU'S Learning Program