Question:
The molecular weight of benzoic acid in benzene as determined by depression in freezing point method corresponds to
The molecular weight of benzoic acid in benzene as determined by depression in freezing point method corresponds to
Updated On: June 02, 2025
- ionisation of benzoic acid
- dimerisation of benzoic acid
- trimerisation of benzoic acid
- solvation of benzoic acid
Hide Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
In benzene, benzoic acid dimerises as : $ \, \, \, \, \, \, C_6H_5COOH \rightleftharpoons \frac{1}{2}(C_6H_5COOH)_2$
Was this answer helpful?
0
0
Top Questions on Colligative Properties
- At \( T \)(K), the vapour pressure of pure benzene and toluene are \( 75 \) and \( 22 \) mm Hg respectively. 23.4 g of benzene and 64.4 g of toluene are mixed to form an ideal solution. If the vapours are in equilibrium with the liquid mixture, the mole fraction of toluene in vapour phase is:
- AP EAPCET - 2025
- Chemistry
- Colligative Properties
- A solution of urea in water has a boiling point of \( 100.18^\circ C \). What is the freezing point of the same solution, if \( K_f \) and \( K_b \) of water are \( 1.86 \) and \( 0.52 \) K kg mol\(^{-1} \), respectively? (Boiling point of water = \( 100^\circ C \))
- AP EAPCET - 2025
- Chemistry
- Colligative Properties
- At 298 K, 0.714 moles of liquid A is dissolved in 5.555 moles of liquid B. The vapour pressure of the resultant solution is 475 torr. The vapour pressure of pure liquid A at the same temperature is 280.7 torr. What is the vapour pressure of pure liquid B in torr?
- AP EAPCET - 2025
- Chemistry
- Colligative Properties
- 1.95 g of non-volatile and non-electrolyte solute dissolved in 100 g of benzene lowered the freezing point of it by 0.64 K. The molar mass of the solute (in g mol\(^{-1}\)) (\(K_f(\text{C}_6\text{H}_6) = 5.12 \, \text{K kg mol}^{-1}\)) is
- AP EAPCET - 2025
- Chemistry
- Colligative Properties
- At \( T \)(K), the vapour pressure of pure benzene and toluene are \( 75 \) and \( 22 \) mm Hg respectively. 23.4 g of benzene and 64.4 g of toluene are mixed to form an ideal solution. If the vapours are in equilibrium with the liquid mixture, the mole fraction of toluene in vapour phase is:
- AP EAPCET - 2025
- Chemistry
- Colligative Properties
View More Questions
Questions Asked in JEE Advanced exam
- At 25$^\circ$C, the concentration of H$^+$ ions in $ 1.00 \times 10^{-3} \, \text{M} $ aqueous solution of a weak monobasic acid having acid dissociation constant $ K_a = 4.00 \times 10^{-11} $ is $ X \times 10^{-7} \, \text{M} $. The value of $ X $ is ____. {Use: Ionic product of water $ K_w = 1.00 \times 10^{-14} $ at 25$^\circ$C
- JEE Advanced - 2025
- Ionic Equilibrium
- Consider the vectors $$ \vec{x} = \hat{i} + 2\hat{j} + 3\hat{k},\quad \vec{y} = 2\hat{i} + 3\hat{j} + \hat{k},\quad \vec{z} = 3\hat{i} + \hat{j} + 2\hat{k}. $$ For two distinct positive real numbers $ \alpha $ and $ \beta $, define $$ \vec{X} = \alpha \vec{x} + \beta \vec{y} - \vec{z},\quad \vec{Y} = \alpha \vec{y} + \beta \vec{z} - \vec{x},\quad \vec{Z} = \alpha \vec{z} + \beta \vec{x} - \vec{y}. $$ If the vectors $ \vec{X}, \vec{Y}, \vec{Z} $ lie in a plane, then the value of $ \alpha + \beta - 3 $ is ________.
- Let $ x_0 $ be the real number such that $ e^{x_0} + x_0 = 0 $. For a given real number $ \alpha $, define $$ g(x) = \frac{3xe^x + 3x - \alpha e^x - \alpha x}{3(e^x + 1)} $$ for all real numbers $ x $. Then which one of the following statements is TRUE?
- JEE Advanced - 2025
- Fundamental Theorem of Calculus
Two identical concave mirrors each of focal length $ f $ are facing each other as shown. A glass slab of thickness $ t $ and refractive index $ n_0 $ is placed equidistant from both mirrors on the principal axis. A monochromatic point source $ S $ is placed at the center of the slab. For the image to be formed on $ S $ itself, which of the following distances between the two mirrors is/are correct:
- JEE Advanced - 2025
- Ray optics and optical instruments
- Let $$ \alpha = \frac{1}{\sin 60^\circ \sin 61^\circ} + \frac{1}{\sin 62^\circ \sin 63^\circ} + \cdots + \frac{1}{\sin 118^\circ \sin 119^\circ}. $$ Then the value of $$ \left( \frac{\csc 1^\circ}{\alpha} \right)^2 $$ is \rule{1cm}{0.15mm}.
- JEE Advanced - 2025
- Some Applications of Trigonometry
View More Questions
JEE Advanced Notification
Concepts Used:
Colligative Properties
Colligative Property of any substance is entirely dependent on the ratio of the number of solute particles to the total number of solvent particles but does not depend on the nature of particles. There are four colligative properties: vapor pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure.
Examples of Colligative Properties
We can notice the colligative properties of arrangements by going through the accompanying examples:
- On the off chance that we add a spot of salt to a glass full of water, its freezing temperature is brought down impressively than its normal temperature. On the other hand, the boiling temperature is likewise increased and the arrangement will have a lower vapor pressure. There are also changes observed in its osmotic pressure.
- In the same way, if we add alcohol to water, the solution’s freezing point goes down below the normal temperature that is usually observed for either pure alcohol or water.
Types of Colligative Properties
- Freezing point depression: ΔTf =1000 x kf x m2 /(M2 x m1)
- Boiling point elevation: ΔTb = kb m
- Osmotic pressure: π = (n2/V) RT
- Relative lowering of vapor pressure: (Po - Ps)/Po