In the context of the lanthanoids, which of the following statement is not correct?
There is a gradual decrease in the radii of the members with increasing atomic number in the series.
All the member exhibit +3 oxidation state.
Because of similar properties, the separation of lanthanoids is not easy.
Because of similar properties, the separation of lanthanoids is not easy.
D.
Because of similar properties, the separation of lanthanoids is not easy.
The freezing point of benzene decreases by 0.45°C when 0.2 g of acetic acid is added to 20 g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be
(Kf for benzene = 5.12 K kg mol–1)
64.6%
80.4%
74.6%
74.6%
D.
74.6%
In benzene
2CH3COOH ⇌ (CH3COOH)2
A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some of the CO2 is converted into CO on the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of K is
1.8 atm
3 atm
0.3 atm
0.3 atm
A.
1.8 atm
Kp depends upon the partial pressure of reactants and products so first calculate their partial pressure and then, calculate Kp,
If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous solution, the change in freezing point of water (ΔTf), when 0.01 mol of sodium sulphate is dissolved in 1 kg of water, is (Kf = 1.86 K kg mol–1).
0.0372 K
0.0558 K
0.0744 K
0.0744 K
B.
0.0558 K
Na2SO4 → 2Na+ + SO42-
Van't Hoff factor of dissociation for Na2SO3 = 3
Molality of the solution = 0.01 m
ΔTf = i x Kf x m
= 3 x 1.86 x 0.01
= 0.0558 K
On mixing, heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components (heptane and octane) are 105 kPa and 45 kPa respectively.Vapour pressure of the solution obtained by mixing 25.0 g of heptane and 35 g of octane will be (molar mass of heptane = 100 g mol–1 and of octane =114 g mol–1)
144.5 kPa
72.0 kPa
36.1 kPa
36.1 kPa
B.
72.0 kPa
Vapour pressure of the solution can be calculated by the formula
pT = XHp°H + Xo.P°o.
Therefore calculate the mole fractions of heptane and octane
pT = XHp°H + Xo.P°o.