PHYSICAL CHEMISTRY
DAILY PRACTICE PROBLEM
TARGET–NEET(UG)
DPP–1
LIQUID SOLUTION (CONCENTRATION TERMS)
1. What is the molarity of SO2–
4
ion in 6. Calculate the molality of a sulphuric
aqueous solution that contain 34.2 ppm acid solution in which the mole fraction
of Al2(SO4)3 (Assume complete of water is 0.85 :
dissociation and density of solution 1 (1) 9.8 m (2) 10.2 m
(3) 9.2 m (4) 10.8 m
g/mL) :
(1) 3 × 10–4 M (2) 2 × 10–4 M
7. What is the mole fraction of a solute in
(3) 10–4 M (4) None of these
2.5 m aqueous solution ?
(1) 55.55 (2) 0.957
2. Molarity and molality of a solution of a
(3) 0.043 (4) 2.7
liquid (mol. wt. = 50) in aqueous
solution is 9 and 10 respectively. What 8. A 6.90 M solution of KOH in water
is the density of solution ? contains 30% by mass of KOH.
(1) 1 g/cc (2) 0.95 g/cc Calculate the density of the KOH
(3) 1.05 g/cc (4) 1.35 g/cc solution.
(Molar mass of KOH = 56 g mol –1) :
3. 0.2 mole of HCl and 0.2 mole of barium (1) 3.86 gm mL–1 (2) 0.23 gm mL–1
chloride were dissolved in water to (3) 2.07 gm mL–1 (4) 1.29 gm mL–1
produce a 500 ml solution. The molarity
9. Molarity of H2SO4 is 0.8 and its density
of the Cl– ions is :
is 1.06 g/cm3. What will be its
(1) 0.06 M (2) 0.09 M
concentration in terms of molality and
(3) 1.2 M (4) 0.80 M
mole fraction ?
(1) 0.785 mol kg–1 & 0.028 respectively
4. Calculate the mole fraction of ethylene (2) 0.815 mol kg–1 & 0.014 respectively
glycol (C2H6O2) in a aqueous solution (3) 0.785 mol kg–1 & 0.014 respectively
containing 20% of C2H6O2 by mass : (4) 0.815 mol kg–1 & 0.028 respectively
(1) 0.932 (2) 0.731
(3) 0.068 (4) 0.621 10. One litre of N/2 HCl solution is heated
in a beaker. It was observed that when
5. Find the molarity of a 15% by mass the volume of the solution was reduced
solution of H2SO4 (density of H2SO4= to 600 ml, 3.25 g of HCl is lost.
Calculate the normality of the new
1.020 g cm–3).
solution:
(1) 1.13 M (2) 1.56 M
(1) 18.25 N (2) 0.685 N
(3) 1.8 M (4) None of these
(3) 0.600 N (4) 3.25 N
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11. What volume of 10% (w/v) solution of 14. The strength of H2O2 (in g/litre) in 11.2
Na2CO3 will be required to neutralise volume solution of H2O2 is :
100 mL of HCl solution containing 3.65 (1) 17 (2) 51
g of HCl ? (3) 34 (4) 85
(1) 212.8 mL (2) 106.4 mL
(3) 53.2 mL (4) 208.2 mL 15. 100 volume hydrogen peroxide solution
means :
12. A solution obtained by mixing 300 g of (1) 17.86 N (2) 30.36% H2O2
25% and 400 g of 40% solution by (3) 8.93 M (4) All are correct
mass. Calculate the mass percentage of
solute in the resulting solution ?
(1) 66.5% (2) 33.5%
(3) 23.4% (4) 76.6%
13. 30 volumes H2O2 means :
(1) 30% H2O2
(2) 30 cm3 of the solution contains 1g
of H2O2
(3) 1 cm3 of the solution liberated 30
cm3 of O2 at STP
(4) 30 cm3 of the solution contain one
mole of H2O2
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PHYSICAL CHEMISTRY
DAILY PRACTICE PROBLEM
TARGET–NEET(UG)
DPP–2
LIQUID SOLUTION (VAPOUR PRESSURE, RAOULT'S LAW)
1. An aqueous solution of glucose is 5. The vapour pressure of benzene at
prepared at 303 K by dissolving 10 g 25ºC is 639.7 mm of Hg and the vapour
glucose in 90 g water. If vapour
pressure of solution of a solute in
pressure of water is 32.8 mm Hg, what
benzene at the same temperature is
is the vapour pressure of solution ?
631.9 mm of mercury. Calculate the
(1) 35.46 mm (2) 29.42 mm
molality of the solution :
(3) 42.74 mm (4) 32.44 mm
(1) 0.0122 (2) 0.158
2. Calculate vapour pressure of a 5% (by (3) 0.9818 (4) 0.844
weight) solution of water in glycerol
(mol. wt. 92.1) at 100ºC, assuming 6. At 40ºC, the vapour pressure (in torr)
Raoult's law to be valid and neglecting of methyl alcohol (1) and ethyl alcohol
the vapour pressure of glycerol : (2) solution is represented by : P = 120
(1) 103.14 mm (2) 127.77 mm XA + 138; where XA is mole fraction of
(3) 162.44 mm (4) 751.33 mm methyl alcohol. The value of lim X A 0
PBo PAo
3. Benzene and toluene form an ideal and lim XB 0 are :
XB XA
solution. The vapour pressure of pure
benzene and toluene are respectively (1) 138, 258 (2) 258, 138
75 mm and 22 mm at 20ºC. If the mole (3) 120, 138 (4) 138, 125
fractions of benzene and toluene in
solution are 0.63 and 0.37 respectively, 7. The vapour pressure in mm (Hg) of a
calculate the vapour pressure of the CH3OH–C2H5OH binary solution p, at a
ideal mixture : certain temperature is represented by
(1) 55.39 mm (2) 119.04 mm the equation : P = 254 – 119 X, where
(3) 178.49 mm (4) None X is the mole fraction of C2H5OH. Find
the vapour pressure of the pure C2H5OH
4. Consider the following vapour pressure
(1) 254 mm (2) 119 mm
composition graph, SP is equal to :
(3) 135 mm (4) 373 mm
sure
r pres
vapou pºB
Tot al P 8. The vapour pressure of a given liquid
Q
Vapour pºA PA will decrease if :
pressure
PB R (1) surface area of liquid is decreased
S
(2) the volume of liquid in the container
0.2 0.4 0.6 0.8 1
mole fraction is decreased
(1) PQ + RS (2) PQ + QR + RS (3) the volume of the vapour phase is
(3) SR + SQ (4) PQ + QR increased
(4) the temperature is decreased
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9. Raoult's law is obeyed by each 12. At 25ºC, the vapour pressure of pure
constituent of a binary liquid solution liquid A (mol. wt. = 40) is 100 torr,
when : while that of pure liquid B is 40 torr,
(1) the forces of attractions between (mol. wt.= 80). The vapour pressure at
like molecules are greater than those 25ºC of a solution containing 20 g of
between unlike molecules each A and B is :
(2) the forces of attraction between like (1) 80 torr (2) 59.8 torr
molecules are smaller than those (3) 68 torr (4) 48 torr
between unlike molecules
(3) the forces of attraction between like 13. Two liquids A and B from ideal
molecules are identical with those solutions. At 300 K, the vapour
between unlike molecules pressure of solution containing 1 mole
(4) the volume occupied by unlike of A and 3 mole of B is 550 mm Hg. At
molecules are different the same temperature, if one more
mole of B is added to this solution, the
10. 6.0 g of urea (molecular weight = 60) vapour pressure of the solution increase
was dissolved in 9.9 moles of water. If by 10 mm Hg. Determine the vapour
the vapour pressure of pure water is Pº, pressure of A and B in their pure states
the vapour pressure of solution is : (in mm Hg) :
(1) 0.10 Pº (2)1.10 Pº (1) 400, 600 (2) 500, 500
(3) 0.90 Pº (4) 0.99 Pº (3) 600, 400 (4) None of these
11. Estimate the lowering of vapour
pressure due to the solute (glucose) in
a 1.0 M aqueous solution at 100ºC :
(1) 100 torr (2) 18 torr
(3) 13.45 torr (4) 24 torr
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PHYSICAL CHEMISTRY
DAILY PRACTICE PROBLEM
TARGET–NEET(UG)
DPP–3
LIQUID SOLUTION (IDEAL and NON IDEAL SOLUTION AZEOTROPIC MIXTURE, HENRY LAW)
1. At what partial pressure, oxygen will 4. Negative deviation from Raoult's law is
–1
have a solubility of 0.05 g L in water observed in which one of the following
at 293 K ? Henry's constant (KH) for O2 binary liquid mixtures ?
in water at 293 K is 34.86 k bar. (1) Ethanol and acetone
Assume the density of the solution to (2) Benzene and toluene
be same as that of the solvent ? (3) Acetone and chloroform
(1) 2.81 × 10–5 bar
(4) Chloroethane and bromoethane
(2) 0.98 bar
(3) 1.32 bar
5. If liquids A and B form an ideal solution,
(4) 1.24 × 10–5 bar
then
(1) enthalpy of mixing is zero
2. Air contains O2 and N2 in the ratio of 1 :
(2) entropy of mixing is zero
4. Calculate the ratio of solubilities in
(3) free energy of mixing is zero
terms of mole fractions of O2 and N2
(4) free energy as well as the entropy
dissolved in water at atmospheric
of mixing are each zero.
pressure and room temperature at
which Henry's constant for O2 and N2
6. In a mixture A and B, components show
are 3.30 × 107 torr and 6.60 × 107 torr
respectively : negative deviation as
(1) 1 : 2 (2) 2 : 1 (3) 1 : 4 (4) 4 : 1 (1) Tb < 0
(2) Vmix < 0
3. Total vapour pressure of mixture of 1 (3) A – B interaction is weaker than A–
mol X Pº
x 150 torr and 2 mol Y
A and B–B interaction
(4) None of the above reason is correct
Pº
Y
300 torr is 240 torr.n this case :
(1) There is a negative deviation from 7. Which of the following gas has highest
Raoult's law value of Henry’s constant (kh) at
(2) There is a positive deviation from constant temperature ?
Raoult's law (1) He (2) CH4
(3) There is no deviation from Raoult's (3) O2 (4) CO2
law
(4) Can not be decided
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PHYSICAL CHEMISTRY
DAILY PRACTICE PROBLEM
TARGET–NEET(UG)
DPP–4
LIQUID SOLUTION (COLLIGATIVE PROPERTY)
1. An aqueous solution of urea is found to
4. At 20ºC, the O.P. of a urea solution is400
boil at 100.94ºC. Given Kb for water is
mm. The solution is diluted and the
–1
0.52ºC molal , what is the mole
temperature is raised to 35ºC, when
fraction of urea in the solution ?
the osmotic pressure is found to be
(1) 0.017 (2) 0.052
105.5 mm. Calculate the extent of
(3) 0.031 (4) 0.042
dilution :
(1) 6 times (2) 2 times
2. Osmotic pressure of a solution obtained
(3) 4 times (4) 8 times
by dissolving 18g of an organic
compound in 1 L solution is 2.414 × 105
5. If boiling point of an aqueous solution is
–2 –1
Nm at 293 K. If value of R is 8.3 JK
100.1ºC. What is its freezing point ?
–1
mol , calculate the molecular mass of
Given latent heat of fusion and
the compound :
vaporization of water are 80 cal g–1 and
(1) 18133 (2) 181.33
540 cal g–1 respectively : (Antilog 0.40
(3) 362.66 (4) 437.74
= 2.57)
(1) –0.181ºC (2) –3.62ºC
3. A solution containing 0.1 g of a non-
(3) –0.362 K (4) –0.362ºC
volatile organic substance P in 100 g of
benzene raises the boiling point of
6. The vapour pressure of a solution of a
benzene by 0.2ºC, while a solution
non-volatile non-electrolyte (A) in a
containing 0.1 g of another non-
solvent (B) is 95% of the vapour pressure
volatile substance Q in the same
of the solvent at the same temperature.
amount of benzene raises the boiling
If MB = 0.3 MA, where MB and MA are
point of benzene by 0.4ºC. What is the
molecular weight of B and A respectively,
ratio of molecular masses of P and Q ?
the weight ratio of the solvent and solute
(1) 1 : 2 (2) 2 : 1
are :
(3) 1 : 4 (4) 4 : 1
(1) –0.17 (2) 5.7
(3) 0.2 (4) 4.0
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7. Which of the following aqueous 10. A 5.8% NaCl solution will exert an
solutions will boil at the highest osmotic pressure closest to that of the
temperature ? following solution
(1) 5.8% sucrose solution
(1) A 0.10 m solution of NaCl
(2) 5.8% glucose solution
(2) A 0.10 m solution of Na2SO4.
(3) 2.0 molal glucose solution
(3) A 0.10 m solution of Na3PO4.
(4) 1.0 molal glucose solution
(4) A 0.10 m solution of K2SO4.
11. Which of the following is the correct
8. In a 0.2 molal aqueous solution of a order of boiling points of the following
weak acid HX, the degree of ionization solutions ?
is 0.3. Taking Kf for water as 1.85, the (I) 0.15 M NaCl
freezing point of the solution will be (II) 0.4 M glucose solution
nearest to (III) 0.15 M CaCl2 solution
(1) – 0.480ºC (2) – 0.360ºC (IV) 0.15 M urea solution
(1) IV<I<III<II (2)IV<II<I< III
(3) – 0.260ºC (4) + 0.480ºC
(3) IV<I<II<III (4) IV<III<II<I
9. Osmotic pressure of a solution is
0.0821 atm at a temperature of 300 K.
The concentration in mol/L will be
(1) 0.33 (2) 0.066
(3) 0.3 × 10–2 (4) 3
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PHYSICAL CHEMISTRY
DAILY PRACTICE PROBLEM
TARGET–NEET(UG)
DPP–5
LIQUID SOLUTION (ABNORMAL COLLIGATIVE PROPERTY)
1. The freezing point of a solution 5. 1.1 g of CoCl3.6NH3 (mol. wt. = 267)
was dissolved in 100 g of H2O. The
containing 0.2 g of acetic acid in 20.0 g
freezing point of the solution was –
benzene is lowered by 0.45ºC. 0.3ºC. How many moles of solute
Calculate the degree of dimerization of particles exist in solution for each mole
acetic acid in benzene, Kf for benzene is of solute introduced ?
Kf for H2O = 1.86ºC.mol–1 :
5.12 K mol–1 kg :
(1) 4 (2) 2 (3) 3 (4) 5
(1) 0.527 (2) 0.80
(3) 0.945 (4) None of these 6. 20.27 g of benzene containing 0.2965 g
of benzoic acid (mol. wt. = 122) freezes
at 0.317º below the freezing point of
2. The vapour pressure of a solution
pure benzene. If benzoic acid exists as
containing 2g of an electrolyte BA in dimer in benzene, find its degree of
100 g water, which dissociates in one association. Kf for benzene is 5.12ºC.m–1
B+ and one A– ion in water, is 751 mm, (1) 51.58% (2) 96.84%
(3) 3.16% (4) 49.42%
at 100ºC. Calculate degree of ionisation
of BA if its mol. mass is 56 : 7. The degree of dissociation of Ca(NO3)2
(1) 86.4 % (2) 24.28 % in a dilute aqueous solution containing
(3) 95.2 % (4) 75.6 % 7 g of the salt per 100 g of water at
100ºC is 70 percent. If the vapour
pressure of water at 100ºC is 760 mm,
3. What will be the osmotic pressure of calculate the vapour pressure of the
0.1M monobasic acid if its pH is 2 at solution :
250ºC ? (1) 746.10 mm (2) 754.20 mm
(3) 750.07 mm (4) None of these
(1) 2.45 atm (2) 2.69 atm
(3) 3.42 atm (4) 4.72 atm 8. Which of the following has been
arranged in order of decreasing freezing
4. A 0.01 m aqueous solution of point :
(1) 0.05 M KNO3 > 0.04 M BaCl2 >
K3[Fe(CN)6] freezes at –0.062ºC. what
0.140 M sucrose > 0.075 M CuSO4
is the apparent percentage of (2) 0.04 M BaCl2 > 0.140 M sucrose >
dissociation ? 0.075 M CuSO4 > 0.05 M KNO3
(Kf for water = 1.86) : (3) 0.075 M CuSO4 > 0.140 M sucrose
> 0.04 M BaCl2 > 0.05 M KNO3
(1) 22% (2) 62%
(4) 0.075 M CuSO4 > 0.05 M KNO3 >
(3) 78% (4) 38% 0.140 M sucrose > 0.04 M BaCl 2
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9. Two solutions S1 and S2 containing 0.1 12. A 0.025 M solution of a monobasic acid
M NaCl(aq) and 0.05 M BaCl 2(aq) are had a freezing point of –0.06ºC.
separated by semipermeable Calculate Ka for the acid. Kf (H2O) =
membrane. Which among the following 1.86 :
statement(s) is/are correct ? (Assume (Assume M = m)
complete dissociation of both the (1) 2.96 × 10–3 (2) 1.2 × 10–3
electrolytes) : (3) 25 × 10–3 (4) 1.5 × 10–3
13. 3.24 g of Hg(NO3)2 (molar mass = 324)
dissolved in 1000 g of water constitutes
a solution having a freezing point of –
0.0558ºC while 21.68 g of HgCl 2 (molar
(1) S1 and S2 are isotonic
mass = 271) in 2000 g of water
(2) S1 is hypertonic while S2 is
constitutes a solution with a freezing
hypotonic
point of –0.0744ºC. The Kf for water is
(3) S1 is hypotonic while S2 is
1.86 K Kg × mol–1. About the state of
hypertonic
ionization of these two solids in water it
(4) All the above
can be inferred that :
(1) Hg(NO3)2 and HgCl2 both are
10. 17.4% (mass/V) K2SO4 solution at 27ºC
completely ionized
is isotonic to 5.85% (mass/V) NaCl
solution at 47ºC. If NaCl is 100% (2) Hg(NO3)2 is fully ionized but HgCl2
ionised, what is % ionisation of K2SO4 is fully unionized
in aqueous solution ? (3) Hg(NO3)2 and HgCl2 both are
(1) 50% (2) 62.5 % completely unionized
(3) 56.5 % (4) 75% (4) Hg(NO3)2 is fully unionized but
HgCl2 is fully ionized
11. 1g of a monobasic acid when dissolved
in 100 g of water lowers the freezing 14. A 0.2 m aqueous solution of KCl freezes
point by 0.168ºC. 0.2 g of the same at –0.68ºC. Calculate the osmotic
acid when dissolved and titrated pressure at 0ºC. Assume the volume of
required 15.1 mL of N/10 alkali. solution to be that of pure H2O and Kf
Calculate the degree of dissociation of for H2O is 1.86 :
the acid. (1) 8.2 atm (2) 4.48 atm
(Kf for water is 1.86) (3) 2.8 atm (4) None of these
(1) 0.19 (2) 0.81
(3) 0.14 (4) 0.84
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