Advertisements
Advertisements
प्रश्न
Why is the value of van't Hoff factor for ethanoic acid in benzene close to 0.5?
Advertisements
उत्तर
In benzene, two molecules of ethanoic acid associate to form a dimer.
\[\ce{2CH3COOH <=> (CH3COOH2)}\]
Thus, the van't Hoff factor,
i = `"Number of solutes after dissociation or associations"/"Number of solutes before dissociation or associations"`
= `1/2`
= 0.5
APPEARS IN
संबंधित प्रश्न
Derive van’t Hoff general solution equation.
The substance ‘X’, when dissolved in solvent water gave molar mass corresponding to the molecular formula ‘X3’. The van’t Hoff factor (i) is _______.
(A) 3
(B) 0.33
(C) 1.3
(D) 1
3.9 g of benzoic acid dissolved in 49 g of benzene shows a depression in freezing point of 1.62 K. Calculate the van't Hoff factor and predict the nature of solute (associated or dissociated).
(Given : Molar mass of benzoic acid = 122 g mol−1, Kf for benzene = 4.9 K kg mol−1)
Define the term abnormal molar mass.
The van’t Hoff factor (i) accounts for ____________.
Van't Hoff factor I is given by expression.
When 9.45 g of ClCH2COOH is added to 500 mL of water, its freezing point drops by 0.5°C. The dissociation constant of ClCH2COOH is x × 10−3. The value of x is ______. (Rounded-off to the nearest integer)
[\[\ce{K_{f(H_2O)}}\] = 1.86 K kg mol−1]
A storage battery contains a solution of H2SO4 38% by weight. At this concentration, Van't Hoff Factor is 2.50. At the battery content freeze temperature will be ______ K.
(Kf = 1.86 K Kg mol−1)
Consider the reaction
\[\begin{bmatrix}\begin{array}{cc}
\phantom{.......}\ce{CH3}\\
\phantom{....}|\\
\ce{CH3CH2CH2 - \overset{⊕}{N} - CH2CH3}\\
\phantom{....}|\\
\phantom{.......}\ce{CH3}
\end{array}\end{bmatrix}\]\[\ce{OH^- ->[Heat] ?}\]
Which of the following is formed in a major amount?
Why is boiling point of 1 M NaCl solution more than that of 1 M glucose solution?
