Advance Ranker DPP 5 Liquid and Atomic
Advance Ranker DPP 5 Liquid and Atomic
Advance Ranker DPP 5 Liquid and Atomic
PRACTICE PROBLEMS
1. x mole of KCI and y mole of BaCl2 are both dissolved in 1 kg of water. Given that x + y = 0.1 and
Kf for water is 1.85 K/molal, what is the observed range of Tf, if the ratio of x to y is varied ?
(A) 0.37 ºC to 0.555 ºC (B) 0.185 ºC to 0.93 ºC
(C) 0.56 ºC to 0.93 ºC (D) 0.37 ºC to 0.93 ºC
2. When only a little quantity of HgCl2(s) is added to excess KI(aq) to obtain a clear solution,
which of the following is true for this solution? (no volume change on mixing).
The reaction is 4KI(aq.) + HgCl2(s) ⎯→ K2[HgI4] (aq.) + 2KCl (aq.)
(A) Its boiling and freezing points remain same
(B) Its boiling point is lowered
(C) Its vapour pressure become lower
(D) Its boiling point is raised
3. The vapour pressure of two miscible liquids (A) and (B) are 300 and 500 mm of Hg
respectively. In a flask, 10 moles of (A) are mixed with 12 moles of (B). However, as soon as
(B) is added, (A) starts polymerizing into a completely insoluble solid. The polymerization
follows first-order kinetics. After 100 minutes, 0.525 moles of a solute is dissolved which
arrests the polymerisation completely. The final vapour pressure of the solution is 400 mm of
Hg. The rate constant of the polymerisation reaction is (Assume negligible volume change on
mixing and polymerisation and ideal behaviour for the final solution:)
4. Tritium, T (an isotope of H ) combines with fluorine to form weak acid TF, which ionizes to
give T + . Tritium is radioactive and is a 𝛽-emitter. A freshly prepared aqueous solution of TF
has pT (equivalent of pH ) of 1.5 and freezes at −0.372∘ C. If 600ml of freshly prepared solution
were allowed to stand for 24.8 years. Calculate (i) ionization constant of TF. (ii) Number of 𝛽-
particles emitted.
(Given K f for water = 1.86 kg mol K −1 , t1/2 for tritium = 12.4 years)
5. Pressure over ideal binary liquid mixture containing 20 moles each of liquid A and B is
gradually decreased isothermally. If PA0 = 200 mm Hg PB0 = 100 mm Hg, find the pressure at
which half of the liquid is converted into vapour.
1
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Liquid Solution & Atomic Structure (Physical Chemistry)
9. The hydrogen like species for which data is given in above passage is -
(A) He+ (B) Li+2 (C) Be+3 (D) B+5
1 P2,1 K
(C) (R) 2 2,3
2 E3,1
E1,2
E v 2,3
(D) 1,4 (S)
E v1,1
2,2
v − v 2,1
(T) 4 2,5
v − v
2,3 2,2
2
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Liquid Solution & Atomic Structure (Physical Chemistry)
(A) 1s (P)
r
(B) 2s (Q)
r2
r
(C) 2p (R)
r2
r
(D) 3p (S)
r
(T)
12. The ratio of the wavelength of a proton & -particle will be 1 : 2 if their
(A) Velocity of proton to velocity of particle is in the ratio 1 : 8
(B) Velocity of proton to velocity of particle is in the ratio 8 : 1
(C) Kinetic energy of proton to Kinetic energy of particle is in the ratio 64 : 1
(D) Kinetic energy of proton to Kinetic energy of particle is in the ratio 16 : 1
3
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Liquid Solution & Atomic Structure (Physical Chemistry)
Answer. Key
1. (A) 2. (B) 3. k 1.004 10−4 min−1
4. (i) K a = 7.3 × 10−3 (ii) 4.55 × 1022 5. (141.4)
6. (B) 7. (D) 8. (C) 9. (A)
10. (A) → (P,Q), (B) → (P,Q), (C) → (R,S), (D) → (T)
11. (A) → (S,P); (B) → (R,T); (C) → (S); (D) → (Q,R)
12. (BD)
Solution
1. (A)
x + y = 0.1
If y = 0 , x = 0.1 (0.1 mole KCl)
Tf = I × Kf m = 2 × 0.1 × 1.85 = 0.37
If x = 0 , y = 0.1 (0.1 mole KCl)
Tf = I × Kf m = 3 × 0.1 × 1.85 = 0.555
2. (B)
4 kJ(aq) + HgCl2 (s) ––→ K2[HgI4](aq) + 2KCl(s)
ni 5 1 (LR) 0 0
nf 1 0 1 2
no. of particles initial = 5 × 2 = 10
no. of particles finally = 1 × 2 + 1 × 3 + 2 × 2 = 9
no. of particles are decreasing
Tb , Tb
3. Initial moles of A = 10
Let the number of moles of A when polymerization is arrested be n.
Moles of B = 12
Moles of solute added = 0.525
Total moles = (n + 12 + 0.525)
= (n + 12.525)
n
xA =
( n + 12.525)
12
xB =
( n + 12.525)
P = PA .XA + PB. XB
n 12
or 400 = 300 × + 500
( n + 12.525) ( n + 12.525)
Solving, we get
n = 9.9
For the first order polymerization,
2.303 A 2.303
k= log. 0 or k = log.
t A t −x
2.303 10
= log.
100 9.9
4
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Liquid Solution & Atomic Structure (Physical Chemistry)
Solving, we get
k 1.004 10−4 min−1
4. (i) K a = 7.3 × 10−3 (ii) 4.55 × 1022
5. (141.4)
Let nB be the mole of B present in 10 mol of the mixture that has been vaporized
n 10 − n B
yB = B ; xB = ;
10 10
10 − n B
PB0
PB0 ·x B nB 10
yB = =
P 10 P
n B PB0 PB0 n B n P0
= − B − B 0 …..(1)
10 P 10P 10 P + PB
& P = PA0 + (PB0 − PA0 ) x B
P − PA0 10 − n B P − PA0
xB = 0 = 0
PB − PA0 10 PB − PA0
nB P0 − P
= 0B 0 .....(2)
10 PB − PA
From (1) and (2)
PB0 PB0 − P
=
PB + PB0 PB0 − PA0
P= PA0 ·PB0 = 200 100
= 141.4 mm Hg Ans.
6. (B)
7. (D)
8. (C)
9. (A)
En2 En1 n2 n1
( n2 − n1 )( n2 − n1 + 1) = 6 & n2 ( n2 − 1 )
= 15 n2 = 6 & n1 = 3.
2 2
No. of lines in Balmer series = 6 – 2 = 4
En2 – En1 – = 4.53 = E6 –E3
Z2 Z2
−13.6 + 13.6 = 4.53
36 9
Z=2
species is He+ ion.
5
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