MyQuestionIcon

1780

You visited us 1780 times! Enjoying our articles? Unlock Full Access!

Depression in Freezing Point

Question 2.32...

Question

Question 2.32

Calculate the depression in the freezing point of water when 10g of $C H_{3} C H_{2} C H C l C O O H$ is added to 250 g of water, $K_{a} = 1.4 \times 10^{- 3}$ ; $K_{f} = 1.86 K k g m o l^{- 1}$

Open in App

Solution

Step I: Calculation of degree of dissociation
Mass of solute = 10g
Molar mass of solute
$(C H_{3} C H_{2} C H C l C O O H) = (12 \times 4) + (1 \times 7) + (35.5) + (16 \times 2) = 48 + 7 + 35.5 + 32 = 122.5 g m o l^{- 1}$
Molal concentration of solution
$= \frac{\frac{M a s s}{M o l a r m a s s}}{M a s s o f s o l v e n t (k g)} = \frac{(10 g)}{(122.5 g m o l^{- 1}) \times (0.25 k g)} = 0.326 m$
If $α$ is the degree of dissociation of $C H_{3} C H_{2} C H C l C O O H$ , then
$\begin{matrix} C H_{3} C H_{2} C H C l C O O H ⇌ & C H_{3} C H C l C O O^{-} & + H^{+} I n i t i a l c o n c . C m o l^{- 1} k g & 0 & 0 A t e q u i l i b r i u m C (1 - α) & C α & C α \end{matrix}$
$∴ K_{a} = \frac{C α \times C α}{C (1 - α)} = C α^{2}$
[Considering $(1 - α) = 1$ or dilute solutions]
$α^{2} = \frac{K_{a}}{C}$ or $α = \sqrt{\frac{K_{a}}{C}}$
So, $α = \sqrt{\frac{1.4 \times 10^{- 3}}{0.326}} = \sqrt{42.9 \times 10^{- 4}} = 6.55 \times 10^{- 2} = 0.065$
Step II: Calculation of van't Hoff factor (f)
$\begin{matrix} C H_{3} C H_{2} C H C l C O O H ⇌ & C H_{3} C H_{2} C H C l C O O^{-} & + H^{+} I n i t i a l n o . o f m o l e s & 1 & 0 & 0 M o l e s a t e q u i l i b r i u m & 1 - α & α & α \end{matrix}$
Total number of moles after dissociation
$= (1 - α) + (α) + (α) = (1 + α)$
Van't Hoff fraction
(i) $= \frac{T o t a l n u m b e r o f m o l e s a f t e r d i s s o c i a t i o n}{N u m b e r o f m o l e s b e f o r e d i s s o c i a t i o n}$
$i = \frac{(1 + α)}{1} = (1 + α)$
$= 1 + 0.065 = 1.065$
Step III: Calculation of depression in freezing point $(Δ T_{f})$
$Δ T_{f} = i K_{f} m = (1.065) \times (1.86 K k g m o l^{- 1}) \times (0.326 m o l k g^{- 1})$
$Δ T_{f} = 0.65 K$

Suggest Corrections

thumbs-up

86

Similar questions

Q. Calculate the depression in the freezing point of water when

10 g

C H_{3} C H_{2} C H C l C O O H

250 g

K a = 1.4 \times 10^{- 3}, K_{f} = 1.86 K k g m o l^{- 1}

Q. Calculate the depression in freezing point of water when

20.0 g

C H_{3} C H_{2} C H C l C O O H

500 g

of water.
[Given:

K_{a} = 1.4 \times 10^{- 3}, K_{f} = 1.86 K k g m o l^{- 1}]

Q. Calculate the amount of

K C l

which must be added to 1 kg of water so that the freezing point is depressed by 2K.

(k_{f}

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

Q. Calculate the freezing point of a solution containing 60g of glucose (Molar mass = 180 g

M o l^{- 1})

in 250 g of water.

(k_{f} o f w a t e r = 1.86 K k g m o l^{- 1})

QUESTION 2.33

19.5 g of $C H_{2} F C O O H$ is dissolved in 500g of water. The depression in the freezing point of water observed in 1.0 C. Calculate the van't Hoff factor and dissociation constant of fluoroacetic acid.
$K_{f} = 1.86 K k g m o l^{- 1}$

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Depression in Freezing Point

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Depression in Freezing Point

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon