The depression in freezing point of water observed for the same amount of acetic acid,trichloroacetic acid,and trifluoroacetic acid increases in the order given above. Explain briefly.
(N/A) The depression in freezing point $(\Delta T_f)$ is a colligative property,which depends on the number of particles (ions) produced in the solution.
Acetic acid $(CH_3COOH)$,trichloroacetic acid $(CCl_3COOH)$,and trifluoroacetic acid $(CF_3COOH)$ ionize to different extents.
The acidity depends on the electron-withdrawing effect of the substituent attached to the carboxylic group.
Fluorine $(F)$ is more electronegative than Chlorine $(Cl)$,which is more electronegative than Hydrogen $(H)$.
Therefore,the electron-withdrawing inductive effect follows the order: $CF_3 > CCl_3 > CH_3$.
This makes the $O-H$ bond in trifluoroacetic acid the most polar,leading to the highest degree of ionization.
Since trifluoroacetic acid produces the maximum number of ions,it shows the greatest depression in freezing point.
Thus,the order of depression in freezing point is: $\text{Acetic acid} < \text{Trichloroacetic acid} < \text{Trifluoroacetic acid}$.
Acetic acid $(CH_3COOH)$,trichloroacetic acid $(CCl_3COOH)$,and trifluoroacetic acid $(CF_3COOH)$ ionize to different extents.
The acidity depends on the electron-withdrawing effect of the substituent attached to the carboxylic group.
Fluorine $(F)$ is more electronegative than Chlorine $(Cl)$,which is more electronegative than Hydrogen $(H)$.
Therefore,the electron-withdrawing inductive effect follows the order: $CF_3 > CCl_3 > CH_3$.
This makes the $O-H$ bond in trifluoroacetic acid the most polar,leading to the highest degree of ionization.
Since trifluoroacetic acid produces the maximum number of ions,it shows the greatest depression in freezing point.
Thus,the order of depression in freezing point is: $\text{Acetic acid} < \text{Trichloroacetic acid} < \text{Trifluoroacetic acid}$.
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Similar Questions
$2.7 \ kg$ of each of water and acetic acid are mixed. The freezing point of the solution will be $-x^{\circ} C$. Consider the acetic acid does not dimerise in water,nor dissociates in water. $x = . . . . . . .$ (nearest integer)
[Given : Molar mass of water $= 18 \ g \ mol^{-1}$,acetic acid $= 60 \ g \ mol^{-1}$]
$K_f \ H_2O = 1.86 \ K \ kg \ mol^{-1}$
$K_f$ acetic acid $= 3.90 \ K \ kg \ mol^{-1}$
Freezing point: $H_2O = 273 \ K$,acetic acid $= 290 \ K$
[Given : Molar mass of water $= 18 \ g \ mol^{-1}$,acetic acid $= 60 \ g \ mol^{-1}$]
$K_f \ H_2O = 1.86 \ K \ kg \ mol^{-1}$
$K_f$ acetic acid $= 3.90 \ K \ kg \ mol^{-1}$
Freezing point: $H_2O = 273 \ K$,acetic acid $= 290 \ K$
Find the freezing point depression of a solution having molality $0.25 \ mol \ kg^{-1}$. $(K_{f} = 4.0 \ K \ kg \ mol^{-1})$ (in $K$)
How much amount of $NaCl$ should be added to $600 \ g$ of water $(\rho=1.00 \ g / mL)$ to decrease the freezing point of water to $-0.2^{\circ} C ?$ ............. $gm$
(The freezing point depression constant for water $=2 \ K \ kg \ mol^{-1}$ )
(The freezing point depression constant for water $=2 \ K \ kg \ mol^{-1}$ )
If $6 \ g$ of solute dissolved in $100 \ g$ of water lowers the freezing point by $0.93 \ K$. What is the molar mass of the solute? $(K_{f} = 1.86 \ K \ kg \ mol^{-1})$
Calculate the freezing point of a solution prepared by dissolving $1.8 \ g$ of glucose $(C_6H_{12}O_6)$ in $500 \ g$ of water. The $K_f$ value for water is $1.86 \ K \ kg \ mol^{-1}$. (in $K$)
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