1. XI(J) CHEMISTRY REVIEW TEST-2
USEFUL DATA
Atomic weights:Al = 27, Mg = 24, Cu = 63.5, Mn = 55, Cl = 35.5, O = 16, H = 1, P= 31,Ag = 108, N = 14,
Li = 7, I = 127, Cr = 52, K=39, S = 32, Na = 23, C = 12, Br = 80, Fe = 56, Ca = 40, Zn = 65.4, Pt = 195,
He = 4, Ba = 137
Q.1 In a mass spectrograph experiment, He+ ion was found to strike at distance of x cm. Calculate
distance where H2O+2 ion would strike if
(a) Potential difference 'V' and magnetic field 'B' are kept constant.
(b) Velocity of ions & magnetic field are constant [2+2]
[Ans. (a) 3x/2 (b) 9x/4]
[Sol.(a) r = 2
qB
Vm
2
 r 
q
m
2
2O
H
He
r
r


=
3
2
 2
2O
H
r  = x
2
3
(b) r =
qB
mu
 r 
q
m
2
2O
H
He
r
r


=
9
4
 2
2O
H
r  = x
4
9
]
Q.2 Two vessels connected by a valve of negligible volume. One container (I) has 2.8 g of N2 &
12.7 g of I2 at temperature T1 (K). The other container (II) is completely evacuated. The container
(I) is heated to T2 (K) while container (II) is maintained at T2/3 (K). The valve is opened. Calculate
masses of N2 in both containers after a very long time. I2 sublimes at T2 (K) and volume of vessel
(I) is half that of vessel (II). Neglect vapour pressure of I2 (s). [4]
[Ans. 0.4 g in (I) & 2.4 g in (II)]
[Sol.
Let x mole of N2 diffuse to vessel II
P = final pressure of N2
P (2V) = xR (T2/3)
P (V) = (0.1 – x) R T2
 2 =
)
x
1
.
0
(
3
x

 0.6 – 6 x = x  x = 0.6/7
i.e. mass = 2.4 g
 II has 2.4 g N2 & I has 0.4 g of N2 ]

2. Q.3 H2O2 + 2KI 

 
 yield
%
40
I2 + 2KOH
H2O2 + 2KMnO4 + 3H2SO4 

 
 yield
%
50
K2SO4 + 2MnSO4 + 3O2 + 4H2O
100 ml of H2O2 sample was divided into two parts. First part was treated with KI and KOH
formed required 200 ml of M/2 H2SO4 for neutralisation. Other part was treated with KMnO4
yielding 6.74 lit. of O2 at STP. Using % yield indicated find volume strength of H2O2 sample
used. [4]
[Ans. 50.4 volumes]
[Sol. mol of H2SO4 = 0.1
mol of KOH = 0.2
 mol of H2O2 =
2
2
.
0
×
4
.
0
1
= 0.25
mol of O2 =
4
.
22
74
.
6
= 0.3
mol of H2O2 =
5
.
0
3
3
.
0

= 0.2
Total mol of H2O2 = 0.45
Molarity of H2O2 =
1
.
0
45
.
0
= 4.5 M
Vol. strength = 11.2 × 4.5 = 50.4 volumes ]
Q.4 Rate of diffusion of ozonized oxygen is 5
4
.
0 times that of pure oxygen. Find
(a) Percentage by mass of ozone in the ozonized sample
(b) Degree of dissociation of oxygen assuming pure O2 in the sample initially. [3+1]
[Ans. 60%, 0.6]
[Sol.
2
O
mix
r
r
=
avg
M
32
= 5
4
.
0

avg
M
32
= 0.8  Mavg = 40 g/mol
O2 l
3
2
O3
t = 0 1
t =  1 – 
3
2


40
32
=
1
)
3
/
(
1 

  = 0.6
3
2
mol. of O
mol. of O
=
3
/
2
1



=
4
.
0
4
.
0
= 1 ;
3
2
mass of O
mass of O
=
48
32
=
3
2
mass % of O3 =
5
3
× 100 = 60% ]

3. Q.5 An open vessel at 27°C is heated. Assuming that volume of vessel remains constant. Calculate
(a) Temperature to which vessel was heated till (3/5)th of air in it has been expelled.
(b) Fraction of molecules escaped out when vessel is heated to 900 K
(c) Temperature at which half of the air escapes out. [1.5+1.5+1]
[Ans. (a) 477°C (b) 2/3 (c) 327°C]
[Sol.(a) 






5
3
1
R
300
PV
=
)
R
(
T
PV
 T = 300 ×
2
5
= 750 K = 477°C
(b) Let x be fraction escaped
R
300
PV
(1 – x) =
R
900
PV
 x = 2/3
(c)
R
300
PV
(1 – 0.5) =
TR
PV
 T = 327°C ]
Q.6 Heat released when 1 mole of acid HX neutralizes 1 mol of base BOH
is 3
30 KJ. To a solution containing HX of molarity
10
M
another
solution containing base BOH was slowly added. A plot between heat
liberated (J) and volume of BOH (ml) added was obtained as shown.
(a) Calculate molarity of base.
(b) Calculate volume of original HX solution [2.5+2.5]
[Ans. (a) 0.011 M (b) 66.66 ml]
[Sol.(a) Let V ml of base be added & M be molarity of base
Moles of base = MV × 10–3
Heat released = MV × 10–3 × 3
30 × 103 J = ( 3
30 MV) J
 Slope = 3
30 M =
3
1
M = 1/90 = 0.011 M
(b) 346.4 = 3
30 ×
10
1
× V
V (where V is vol. of HX in ml)
V = 66.66 ml ]
Q.7 200 g of an oleum sample was dissolved in excess water. To this solution 1 kg of impure Ba(NO3)2
was added causing formation of BaSO4 precipitate. The precipitate formed was filtered, washed
& dried. Its weight was 582.5 gm. Find % labelling of oleum sample. [5]
[Ans. 122.5%]
[Sol. moles of BaSO4 = moles of H2SO4 = 2.5 moles
Let x be % labelling of oleum
moles of SO3 = 




 
18
100
x
×2

4. moles of H2SO4 = 












 
 2
80
18
100
x
200
98
1
 2 ×
18
100
x 
+
98
200
– 










 
49
80
18
100
x
= 2.5
 x = 122.5% ]
Q.8 400 ml of 0.2 × 10–3 M Ca(NO3)2 solution is mixed with 400 ml of 10–4 M NaHCO3 solution. If
density of resulting solution is 1 g/ml. Calculate
(a) ppm of Ca(HCO3)2 in final solution
(b) Molality of Na+ in solution
(c) Molarity of Ca2+ in solution [2+2+1]
[Ans. (a) 4.05 ppm (b) 5 × 10–5 mol/kg (c) 10–4 M]
[Sol. Ca(NO3)2 (aq) + 2NaHCO3 (aq)  Ca(HCO3)2 (aq) + 2NaNO3 (aq)
t = 0 80 × 10–6 mol 40 × 10–6 mol
t =  60 × 10–6 mol 0 20 × 10–6 mol 40 × 10–6 mol
Mass of final solution = 1 × 800 = 800 gm
(a) ppm of Ca(HCO3)2 =
800
162
10
20 6

 
× 106 = 4.05 ppm
(b) Molality of Na+ =
800
10
40 6


× 1000 = 5 × 10–5 mol/kg
(c) Molarity of Ca2+ = 3
6
10
800
10
80




= 10–4 M ]
Q.9 A bulb of constant volume is attached to a manometer tube open at other
end as shown in figure. The manometer is filled with a liquid of density
(1/3)rd that of mercury. Initially h was 228 cm. Through a small hole in the
bulb gas leaked causing pressure decrease as
dt
dp
= – kP. If value of h is
114 cm after 7 minutes. Calculate value of k in units of hour–1.
[Use: ln (4/3) = 0.28 & density of Hg = 13.6 g/ml] [5]
[Ans. 2.4 hr–1]
[Sol. P = P0e–kt
P0 =
3
228
+ 76 = 152 cm Hg
At t = 7 min.
P =
3
114
+ 76 = 114 cm Hg
 152e–kt = 114
 k =
t
1
ln
114
152
=
7
1
× 0.28 = 0.04 min–1 = 0.04 × 60 hr–1 = 2.4 hr–1 ]

5. Q.10 Calculate minimum number of balloons each of volume 82.1 lit. required to lift a mass of 1 kg to
a height of 831 m. Given: molar mass of air = 29 g/mol, temperature is constant at 290 K and
mass of each balloon is 50 g. [Use e–0.1 = 0.9, pressure at sea level = 1 atm, acceleration due to
gravity (g) = 10 m/s2] [5]
[Ans. 25 balloons]
[Sol. Density of air at sea level, d0 =
290
0821
.
0
1
29


=
821
.
0
1
g/lit
Density at 831 m = d = d0e–Mgh/RT =
821
.
0
1
×











 
290
31
.
8
831
10
10
29
–
3
e =
821
.
0
1
× e–0.1 =
821
.
0
9
.
0
g/lit
Let n be number of balloons
 103 × g + n × 50g =
821
.
0
9
.
0
× 82.1 × g × n
 1000 + 50 n = 90 n
 n = 25 balloons ]
Q.11 A mixture 1,2–Dipropene and hydrogen gas was placed in a rigid steel container at a constant
temperature of 18°C. Initial pressure of mixture was 10 atm. Sparking the mixture caused
hydrogenation reaction
C3H4 (g) + 2H2(g) C3H8 (g)
causing pressure to decrease by 6 atm. Excess O2 was then added and on sparking the pressure
further decreased by11 atm. Calculate decrease in pressure by treating final mixture with aqueous
KOH solution. [6]
[Ans. 12 atm]
[Sol. If C3H4 is limiting reagent
C3H4 + 2H2 C3H8
t = 0 P 10–P
t =  0 10–3P P
10 – 3P + P = 4  P = 3 atm
On sparking,
C3H8 + 5O2  3CO2 + 4H2O
3 atm 15 atm 9atm
H2 +
2
1
O2  H2O
1atm
2
1
atm
Pressure decrease = (3 + 15 – 9) + (1 + 0.5) = 10.5 atm  11 atm
Case II: H2 is limiting reagent
C3H4 + 2H2 C3H8
t = 0 P 10–P
t =  P–
2
P
10 
0
2
P
10 
 P = 4 atm

6. C3H4 + 4O2  3CO2 + 2H2O
1 atm 4atm 3 atm
C3H8 + 5O2  3CO2 + 4H2O
3 atm 15atm 9 atm
 (1 + 4 – 3) + (3 + 15 – 9) = 11 atm
Pressure of CO2 = 12 atm
 Pressure decreased with KOH solution = 12 atm ]
Q.12 0.2 gm of an anhydrous organic acid gave on combustion 0.04 g water and 0.195 g of CO2. 0.5 g
of its silver salt leaves on ignition 0.355 g silver. In another experiment 100 ml of 0.2 M acid
solution was neutralised by 400 ml of
10
1
M caustic potash solution. What is molecular formula
of acid? [6]
[Ans. C2H2O4]
[Sol. moles of H = 2
18
04
.
0

moles of C =
44
195
.
0
moles of O =
16
1





 


44
195
.
0
12
9
04
.
0
2
.
0
C : H : O = 0.0044 : 0.0044 : 0.0089 = 1 : 1 : 2
Empirical formula = CHO2
Let n be basicity of acid
100 × 0.2 × n = 400 ×
10
1
 n = 2
Let H2X be acid and Ag2X its silver salt
 Ag2X 
 2Ag
0.5 g 0.355
2
M
216
2
5
.
0



=
108
355
.
0
M = Mol. wt. of acid
214
M
1

=
108
355
.
0
M = 90.2
n =
45
2
.
90
 2
Molecular Formula = C2H2O4 ]

7. Q.13 A vessel of volume 246 lit. contains 1.8 g of water and N2 gas at a total pressure of 0.41 atm. If
half the contents of vessel are transferred to another evacuated vessel of volume 24.6 lit. Calculate
(a) Final total pressure in both vessels
(b) Relative humidity in both vessels.
[Assume constant temperature of 300 K. Aqueous tension of water is 15.2 mmHg. ] [3+3]
[Ans. In 246 lit vessel : T
P = 0.205 atm and RH = 25%
In 24.6 lit vessel : T
P = 2.02 atm and RH = 100% ]
[Sol. Moles of H2O = 0.1 mol
Partial pressure of H2O =
246
300
082
.
0
1
.
0 

= 0.01 atm
Aqueous tension of water =
760
2
.
15
= 0.02 atm
 All water is in vapour form
moles of N2 in vessel =
6
.
24
246
4
.
0 
= 4 moles
In each vessel there are 2 moles of N2 & 0.05 moles of water vapour.
In 246 lit. vessel total pressure
P =
246
6
.
24
05
.
2 
= 0.205 atm
In 24.6 lit vessel, partial pressure of water =
6
.
24
6
.
24
05
.
0 
= 0.05 atm
Since this pressure is greater than aqueous tension so, partial pressure of water = 0.02 atm(rest
water in liquid form)
Pressure exerted by N2 =
6
.
24
6
.
24
2
= 2 atm
Total pressure = 2.02 atm
RH of 2nd vessel = 100%
RH of 1st vessel =
02
.
0
005
.
0
× 100 = 25% ]
Q.14 A tube of uniform cross-section of 0.5 cm2 and length 118 cm is divided into three compartments
by using two frictionless pistons each of mass 101.32 gm. The compartments I, II & III are filled
by equal masses of CH4, O2 and He gases respectively as shown. The tube is then inverted at
angle of 30° with horizontal as shown. If pressure in CH4 compartment was 1 atm find volumes
of compartments I, II and III. Assume volume of piston is negligible and temperature is constant.
[Use g = 10 m/s2] [6]
[Ans.VI = 9 ml, VII = 5 ml, VIII = 45 ml]
[Sol. Let P1, P2 & P3 be pressures in I, II & III compartments in atm units

8. Then, P2 – P3 = 3
4
3
10
32
.
101
10
5
.
0
)
2
/
1
(
10
10
32
.
101








atm = 0.1
P1 – P2 = 0.1
Q P1 = 1 atm  P2 = 0.9 atm & P3 = 0.8 atm
Let w gm of each gas be present
1 × V1 =
16
w
RT
T
0.9 × V2 =
32
w
RT
T
0.8 × V3 =
4
w
RT
T
where V1, V2, V3 = Vol. of compartment 1, 2 & 3
V1 : V2 : V3 =
16
1
:
8
.
28
1
:
2
.
3
1
= 1.8 : 1 : 9
V1 =
8
.
11
8
.
1
× 59 ml = 9 ml
V2 =
8
.
11
1
× 59 = 5 ml
V3 =
8
.
11
9
× 59 = 45 ml ]
Q.15 1.8 g of MBrx (M is a metal) when heated with a stream of HCl gas was completely converted to
chloride MClx which weighed 1 g. Specific heat of metal is 0.11 cal/g. Calculate percent error in
molecular weight of metal bromide as determined from specific heat data. [6]
[Ans. 0.7 %]
[Sol. MBrx + xHCl  MClx + xHBr
Atomic mass of metal =
11
.
0
4
.
6
= 58.2

x
80
2
.
58
8
.
1

=
x
5
.
35
2
.
58
1

 0.8(58.2) = (80 – 35.5 × 1.8)x
 x = 2.89  x = 3
Let M be exact atomic wt. of metal
)
3
(
80
M
8
.
1

=
)
3
(
5
.
35
M
1

 M =
8
.
0
3
)
8
.
1
5
.
35
80
( 


 M = 60.3
Exact mol. wt. of MBr3 = 60.3 + 240 = 300.3
Mol. wt. of MBr3 from specific heat data = 58.2 + 240 = 298.2
 % error =
37
.
300
2
.
298
37
.
300 
 0.7% Ans. ]