Advertisements
Advertisements
Question
19.5 g of CH2FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.0°C. Calculate the van’t Hoff factor and dissociation constant of fluoroacetic acid.
Advertisements
Solution
It is given that:
w1 = 500 g
w2 = 19.5 g
Kf = 1.86 K kg mol−1
ΔTf = 1.0°C
We know that:
M2 = `(K_f xx w_2 xx 1000)/(Delta T_f xx w_1)`
= `(1.86 K kg "mol"^(-1) xx 19.5 g xx 1000 g kg^(-1))/(500 g xx 1.0)`
= 72.54 g mol−1
∴ Observed molar mass of CH2FCOOH, (M2)obs = 72.54 g mol−1
The calculated molar mass of CH2FCOOH is (M2)cal = 14 + 19 + 12 + 16 + 16 + 1
= 78 g mol−1
∴ Van’t Hoff factor (i) = `((M_2)_(cal))/(M_2)_(obs)`
= `78/72.54`
= 1.0753
Let α be the degree of dissociation of CH2FCOOH.
| \[\ce{CH2FCOOH ⇌ CH2FCOO^- + H^+}\] | |||
| Initial Conc. | C mol L−1 0 0 | ||
| At equilibrium | C(1− α) Cα Cα | ||
∴ i = `(C(1 + α))/C`
⇒ i = 1 + α
⇒ α = i − 1
= 1.0753 − 1
= 0.0753
Now, the value of Kα is given as:
Kα = `([CH_2FCOO^-][H^+])/([CH_2FCOOH])`
= `(C alpha * C α)/(C (1 - alpha))`
= `(C alpha^2)/(1 - alpha)`
Taking the volume of the solution as 500 mL, we have the concentration:
C = `19.58/78 xx 1/500 xx 1000`
= 0.5 M
∴ Kα = `(C alpha^2)/(1 - alpha)`
= `(0.5 xx (0.0753)^2)/(1 - 0.0753)`
= `(0.5 xx 0.00567)/0.9247`
= 0.00307 (approximately)
= 3.07 × 10−3
RELATED QUESTIONS
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
Define the term abnormal molar mass.
How van’t Hoff factor is related to the degree of dissociation?
How will you convert the following in not more than two steps:
Acetophenone to Benzoic acid
The van’t Hoff factor (i) accounts for ____________.
We have three aqueous solutions of NaCl labelled as ‘A’, ‘B’ and ‘C’ with concentrations 0.1 M, 0.01 M and 0.001 M, respectively. The value of van’t Hoff factor for these solutions will be in the order ______.
The values of Van’t Hoff factors for KCl, NaCl and K2SO4, respectively, are ______.
Van’t Hoff factor i is given by the expression:
(i) i = `"Normal molar mass"/"Abnormal molar mass"`
(ii) i = `"Abnormal molar mass"/"Normal molar mass"`
(iii) i = `"Observed colligative property"/"Calculated colligative property"`
(iv) i = `"Calculated colligative property"/"Observed colligative property"`
Van't Hoff factor I is given by expression.
Geraniol, a volatile organic compound, is a component of rose oil. The density of the vapour is 0.46 g L–1 at 257°C and 100 mm Hg. The molar mass of geraniol is ______ g mol–1. (Nearest Integer)
[Given: R = 0.082 L atm K–1 mol–1]
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)
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 boiling point of 1 M NaCl solution more than that of 1 M glucose solution?
Calculate Van't Hoff factor for an aqueous solution of K3 [Fe(CN)6] if the degree of dissociation (α) is 0.852. What will be boiling point of this solution if its concentration is 1 molal? (Kb = 0.52 K kg/mol)
