Calculate the depression in freezing point of water when $20.0 g$ of $C H_{3} C H_{2} C H C l C O O H$ is added to $500 g$ of water.
[Given: $K_{a} = 1.4 \times 10^{- 3}, K_{f} = 1.86 K k g m o l^{- 1}]$ .

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Solution

∆ $T_{f}$ = $\frac{K_{f} \times W_{2}}{M_{2} \times W_{1}}$
$W_{2}$ = mass of solute = 20g
$M_{2}$ = molar mass of solute = $124.5 g m o l^{- 1}$
$W_{1}$ = Mass of solvent = $500 g$
∆ $T_{f}$ = $\frac{1.86 \times 10^{3} \times 20}{124.5 \times 500}$

∆ $T_{f}$ = $5.98 K$

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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

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}

]

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

Q. (i) The depression in freezing point of water observed for the same molar concentration of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order as stated above. Explain.
(ii) Calculate the depression in freezing point of water when 20.0 g of

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

is added to 500g of water.

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}

for water

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

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Calculate the depression in freezing point of water when 20.0 g of CH3CH2CHClCOOH is added to 500 g of water.[Given: Ka=1.4×10−3,Kf=1.86 K kg mol−1].

∆Tf = Kf×W2M2×W1W2 = mass of solute = 20gM2 = molar mass of solute = 124.5gmol−1W1 = Mass of solvent = 500g∆Tf = 1.86×103×20124.5×500∆Tf = 5.98K

Calculate the depression in freezing point of water when $20.0 g$ of $C H_{3} C H_{2} C H C l C O O H$ is added to $500 g$ of water.
[Given: $K_{a} = 1.4 \times 10^{- 3}, K_{f} = 1.86 K k g m o l^{- 1}]$ .

∆ $T_{f}$ = $\frac{K_{f} \times W_{2}}{M_{2} \times W_{1}}$
$W_{2}$ = mass of solute = 20g
$M_{2}$ = molar mass of solute = $124.5 g m o l^{- 1}$
$W_{1}$ = Mass of solvent = $500 g$
∆ $T_{f}$ = $\frac{1.86 \times 10^{3} \times 20}{124.5 \times 500}$

∆ $T_{f}$ = $5.98 K$