Advertisements
Advertisements
प्रश्न
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)
Advertisements
उत्तर
We know that the depression in freezing point is given by
`DeltaT_`
Here,
van't Hoff factor = i
Depression in freezing point, ΔTf=1.62 K
Kf for benzene=4.9 K kg mol−1
Mass of benzoic acid, ws=3.9 g
Mass of benzene, W=49 g
Molar mass of benzoic acid, Ms=122 g mol−1
Substituting the values, we get
`1.62 = (ixx4.9xx(3.9xx1000))/(122xx49)`
`rArri=(1.62xx122xx49)/(4.9xx3.9xx1000)`
= 0.51
As the value of i < 1, benzoic acid is an associated solute.
APPEARS IN
संबंधित प्रश्न
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
Define van’t Hoff factor.
How van’t Hoff factor is related to the degree of dissociation?
The Van't Hoff factor (i) for a dilute aqueous solution of the strong elecrolyte barium hydroxide is (NEET) ______.
The van’t Hoff factor (i) accounts for ____________.
The values of Van’t Hoff factors for KCl, NaCl and K2SO4, respectively, are ______.
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]
The degree of dissociation of Ca(NO3)2 in a dilute aqueous solution containing 7 g of the salt per 100 g of water at 100°C is 70%. If the vapour pressure of water at 100°C is 760 mm. The vapour pressure of the solution is ______ mm.
When 19.5 g of F – CH2 – COOH (Molar mass = 78 g mol−1), is dissolved in 500 g of water, the depression in freezing point is observed to be 1°C. Calculate the degree of dissociation of F – CH2 – COOH.
[Given: Kf for water = 1.86 K kg mol−1]
Why is the value of van't Hoff factor for ethanoic acid in benzene close to 0.5?
