CHEMISTRY-24-09- 11th (J-Batch) SOLUTION
SOLUTIONS (CHEMISTRY) RTP-3 Calculate rms velocity of dust particles if mass of one particle of dust 8.31×10–12 g at 329.3°C assuming kinetic theory of gases is applicable. [3]
[Sol: m = 8.31 x 10-12g
or 8.31 x 10-12 x 10-3 = 8.31 x 10-15 kg T = 329.3 x 273 = 6.2.3k
Urms =
= 1.732 x 10-3 m/sec.]
Q.1 7 g of an organic compound on treatment with NaOH gave NH3 which when passed through
1
200 ml of 2 M H2SO4 completely neutralized it. Calculate % of nitrogen in the organic compound.
2NH3 + H2SO4 (NH4)2SO4
[Sol: Org. comp + NaOH NH3 H2 SO4 (NH4)2SO4
[3]
2NH3 + H2SO4 (NH4)2SO4
m. moles of H2SO4 used = 200 x 1/2 = 100
m moles of NH3 produced = 100 x 2 = 200
nNH3 = 0.2
nN = 0.2
wN = 0.2 x 14 = 2.8 gm
% N =
2.8 x100 = 40%]
7
Q.2 A tube of uniform cross-section of length 100 cm is divided into two parts by a weightless and frictionless piston. One part contains 4 moles of hydrogen at 2 atm at equilibrium and other part contains 1 mole of nitrogen at the same temperature. Assume volume of piston to be negligible.
(a) Calculate the length of each compartment if the tube was placed horizontally.
(b) The tube is then held at angle of 45° with the horizontal keeping the nitrogen end upwards. Find the length of each compartment. [1.5+1.5]
[Sol:
Let the area of cross section = A
PV1 = PV2 ;
V1 = V2
= A (100 a) = a A
n1T
n2T
n1 n2 4 1
4a = 100 - a
5a = 100
a = 20 cm
(a) Length of H2 compartment = 80 cm; length of N2 compartment = 20 cm.
(b) Unchanged, remain same]
Q.4(a) Find molality of Ca2+ and NO in 2 M Ca(NO )
aqueous solution of density 1.328 g/mL.
3 3 2
(b) Also find mole fraction of solvent in solution. [3+2]
[Sol: Let volume of solution = 1000ml
W solution = 1000 x 1.328 = 1328 g moles of Ca(NO3)2 = 2
WCa( NO3 )2
= 2 x 164 g = 328 g
mass of water = 1000g
(a) molality of Ca2+ = 328 /164
1
molality of NO – = 2 x124 / 62
= 2 moles
= 4 moles
3 1
55.56
(b) mole fraction of H2O = 6 + 55.56
= 0.901]
Q.5 A 7.38 L vessel containing N2 at 1 atm and 27°C is connected to a 8.2 L vessel containing N2 at 4 atm and 127°C thourgh a stopcock. When stopcock is opened, keeping both vessels at their original temperatures, calculate
(a) Final pressure
(b) Moles of N2 in each vessel. [5]
[Sol: (a) (n1 + n2) initially = (n1 + n2) finally
P1V1 + P2V2 = PV1 + PV2
T1 T2 T1 T2
1x 7.38 + 4 x 8.2
= P 7.38 + 8.2
300
400
300
400
0.0246 + 0.082 = P (0.0246 + 0.0205)
0.1066 = P x 0.0451
P = 2.36 atm
(b) In 1st Vessel
In 2nd vessel
nN2 =
PV1 =
RT1
2.36 x 7.38
0.082 x 300
= 0.708
n = PV2 = 2.36 x 8.2
= 0.59]
N2 RT2
0.082 x 400
Q.6 A common method of working of temperature scales is byassigning temperatures for steam point (Boiling point of water) and absolute zero. A temperature scale commonly used by engineers is the Rankine (R) scale on which absolute zero is 0 R and steam point is 671.4 R. Taking linear variation in temperatur scales.
(a) Calculate ice point (melting point of water) on Rankine scale.
(b) Express 1800 R in Kelvin units. [5]
[Ans: (a) 491.4R (b) 1000K]
Q.7 A container holds 3L of N2(g) & liquid water at 300 K at a pressure of 1 atm. The water in container is
1
instantaneously split into H2 & O2 by electrolysis : H2O(l) H2(g) + 2 O2(g). After the completion
of reaction, the total pressure was 1.698 atm. Calculate mass of water taken initially in the container. Given : Aqueous tension of water at 300 K is 0.04 atm. [6]
[Sol:
PN2 + PH2 = 1atm
PN2 = 1 0.04 = 0.96 atm
H O(l) H (g) + 1 O
(g )
2 2 2 2
P
On completion P 2
after the completion of reaction
P
P + 2 + (0.96 + 0.04) = 1.698
3P 1.396
2 = 0.698; P = 3
1.396 x 3
atm
nH2
formed =
3 x 0.082 x 300
= 0.0567
moles of liquid water = moles of H2 produced = 0.0567
mass of liquid water = 18 x 0.0567 = 1.0206g & for gaseous water
PV = nRT
0.04 x 3 = n x 0.082 x 300
n = 4.878 x 10-3
mass of water vapour = 0.0878
total mass of water taken initially in the container = 1.0206 + 0.0878
= 1.1084 = 1.11gm]
Q.8 Compressbility factor Z vs P plot for four real gases A, B, C & D are shown at 300 K
(a) Out of gases A and B whose molecules are larger? Justify.
(b) If slope of curve for B is 0.02 atm–1, calculate Van der Waals constant b for gas B.
(c) Arrange Boyle temperatures (T) for gases A, B, C & D in increasing order. [2+2+2]
[Sol: (a) Gas A have larger molecular size because it has larger value of Z then B.
(b) Z = 1+ bP
RT
dZ = b
dP RT
= 0.02
b = 0.02 x 0.082 x 300 = 0.492
(c) A < B < C < D]
Q.9 Calculate (in terms of NA)
(a) Number of nitrogen atoms in 160 amu of NH4NO3.
(b) Number of gram-atoms of S in 490 kg H2SO4.
(c) Grams of Al2(SO4)3 containing 3.2 amu of S. [6]
[Sol: (a) NH NO = 160 amu =
160
160
NA gm
2
nNH NO = N
80 = N
4 3 A A
molecules of NH4NO3 = 2
atoms of N = 2 × 2 = 4
(b)
H SO
= 490 Kg = 490 × 103 gm
490 103
nH SO
= 98
= 5000
ns = 5000
3.2 1
(c) ns =
NA 32
= 10NA
1
nAl (SO )
= 30N
2 4 3 A
1 114
= × 342 = gm ]
Al2 (SO4 )3
30NA NA
Q.10 20 g mixture of Na2CO3, NaHCO3 and NaCl on heating to moderate temperatures produced 1.12 L of CO2 at NTP. 8 g of the same mixture in a different experiment required 80 ml of 1 M HNO3 for neutralization. Calculate mass % of each component in mixture. [8]
2NaHCO3
Na2CO3 + H2O + CO2
Na2CO3 + 2HNO3 2NaNO3 + H2O + CO2 NaHCO3 + HNO3 NaNO3 + H2O + CO2
[Sol: Let moles of NaHCO3 in 20 gm mixture = b
2NaHCO3 Na2CO3 + H2O + CO2
nCO
nNaHCO b
= 3 =
2 2
b
2 =
1.12
22.4
= 0.05
b = 0.1
Now in 8 gm mixture
nNaHCO =
b
20 × 8 = 0.4 b
& moles of Na2CO3 = a
Na2CO3 + 2HNO3 2NaNO3 + H2O + CO2
a
& NaHCO3 + HNO3 NaNO3 + H2O + CO2
0.4 b
moles of HNO3 required = 2a + 0.4 b
80 1
1000
= 2a + 0.4 b
2a + 0.4 (0.1) = 0.08
a = 0.02
In 8 gm mixture
Na CO
= 0.02 × 106 = 2.12 gm
2 3
NaHCO
= (0.4 × 0.1) × 84 =3.36 gm
% Na2CO3 =
2.12 100
8
= 26.5%
% NaHCO3 =
3.36 100
8
= 42%
& % NaCl =100 – (26.5 + 42) = 31.5% ]
Q.11 A gaseous mixture containing N2 & H2 in mole ratio 1 : 3 is used to carry out the reaction in a sealed container.
N2 (g) + 3H2(g) 2NH3 (g)
The total pressure was 0.8 atm at 25°C before the reaction. Determine final pressure at 323°C assuming
60% dissociation of reaction. [8]
[Sol: Let
P = P & P
2 2
= 3P (initially)
P + 3P = 0.8 atm P = 0.2 atm
P = 0.2 atm &
2
P = 0.6 atm
2
N2(g) + 3H2(g) 2NH3(g) t = 0 0.2 0.6
't' 0.2(1 – 0.6) 0.6 (1 – 0.6) 2(0.2 × 0.6)
= 0.08 = 0.24 = 0.24
Total pressure = 0.08 + 0.24 + 0.24 = 0.56 atm
Now
P1 = P2
T1 T2
0.56 =
298
P2
596
P2 = 1.12 atm ]
Q.12 Calculate compressibility factor (Z) for 0.02 moles of a Van der Waals gas at pressure of 0.1 atm. Assume the size of gas molecules is negligible.
Given:
RT = 20 L-atm-mol–1 (T = Temperature of gas), a = 1000 atm-L2-mol–2 [8]
n2 a
[Sol:
P +
V 2
(V – nb) = nRT
V – nb ~ V
n2 a
P +
V
n2a
V = nRT
PV + V = nRT
(0.02)2 (1000)
0.1 V +
0.1 V +
0.4
V
V
= 0.4
= 0.02 (20)
0.1 V2 – 0.4 V + 0.4 = 0
V2 – 4V + 4 = 0
(V – 2)2 = 0
V = 2 lt.
Z =
PV nRT =
0.1 2
0.02 20
= 0.5 ]
Q.13 Three ideal gas samples in separate equal volume containers are taken and following data is given:
Pressure Temperature Mean free paths Mol. wt.
Gas A 1 atm 1600 K 0.16 nm 20
Gas B 2 atm 200 K 0.16 nm 40
Gas C 4 atm 400 K 0.04 nm 80
Calculate ratio (A : B : C) of following for the three gases.
(a) Average velocities.
(b) Number of collision made by one molecule per sec. (Z1).
(c) Collision frequencies (Z11). [3+3+3]
[Sol: A =
A2 =
1
=
kTA
kTA
kTB
Similarly 2 =
& C2 =
B
kTC
Now (Vavg)A = =
(Vavg)B = =
(Vavg)C = =
= 200
= 200
= 800
Now NA* =
PA
kTA
PB
1
= 1600k
2 1
*
B kTB
PC
= 200 k
4
= 100k
1
NC* =
kTC
= 400 k
= 100k
2 (V
) N *2
kTA
800 R 1 2
A avg A A
(a) (Z11)A =
kT
200
=
R 1 2
2 1600k
(Z )
= B
11 B
2B PB 100k
kT 200 R 1 2
(Z )
= C
11 C
2C PC
100k
(Z11)A : (Z11)B : (Z11)C
kT 800 R
1 2 kT 200 R 1
2 kT 200 R 1 2
= A
: B
: C
2APA
1600k
2B PB
2 100k
2C PC
2 100k
TA
800
TB
200
TC
200
= P
(1600)2 : P
(100)2 : P (100)2
A A
B B
C C
1600 800 200 200 400 200
= 0.16 1 (1600)2 : 0.16 1 (100)2 : 0.04 4 (100)2
16 8
2 2
4 2
= 0.16 (16)2 : 0.32 : 0.16
= 25 : 25 : 50
8 2
= 1 : 1 : 2
8 2
= 1 : 4 : 16 Ans.
2 *
1
(b) (Z1)A = 2 A (Vavg ) NA =
800
A
1
1600k
(Z1)B =
200
100k
1
(Z1)C =
200
100k
TA
800 TB
200 TC
200
(Z1)A : (Z1)B : (Z1)C = P
:
1600 P
:
100 P
100
A A
B
B
C
C
1600 1 :
200
(2):
400
(2)
= 0.16 0.16 2
0.04 4
1 2
= 50 : 12.5 : 50
= 4 : 1 : 4
(c) (Vavg)A : (Vavg)B :(Vavg)C
= 800
= 4 : 1 : 1
: 200
: 200
SOLUTIONS (CHEMISTRY) RTP-3 ACME
Q.1 Calculate rms velocity of dust particles if mass of one particle of dust 8.31×10–12 g at 329.3°C assuming kinetic theory of gases is applicable. [3]
[Sol: m = 8.31 x 10-12g
or 8.31 x 10-12 x 10-3 = 8.31 x 10-15 kg T = 329.3 x 273 = 6.2.3k
Urms =
= 1.732 x 10-3 m/sec.]
Q.2 7 g of an organic compound on treatment with NaOH gave NH3 which when passed through
1
200 ml of 2 M H2SO4 completely neutralized it. Calculate % of nitrogen in the organic compound.
2NH3 + H2SO4 (NH4)2SO4
[Sol: Org. comp + NaOH NH3 H2 SO4 (NH4)2SO4
[3]
2NH3 + H2SO4 (NH4)2SO4
m. moles of H2SO4 used = 200 x 1/2 = 100
m moles of NH3 produced = 100 x 2 = 200
nNH3 = 0.2
nN = 0.2
wN = 0.2 x 14 = 2.8 gm
% N =
2.8 x100 = 40%]
7
Q.3 A tube of uniform cross-section of length 100 cm is divided into two parts by a weightless and frictionless piston. One part contains 4 moles of hydrogen at 2 atm at equilibrium and other part contains 1 mole of nitrogen at the same temperature. Assume volume of piston to be negligible.
(a) Calculate the length of each compartment if the tube was placed horizontally.
(b) The tube is then held at angle of 45° with the horizontal keeping the nitrogen end upwards. Find the length of each compartment. [1.5+1.5]
[Sol:
Let the area of cross section = A
PV1 = PV2 ;
V1 = V2
= A (100 a) = a A
n1T
n2T
n1 n2 4 1
4a = 100 - a
5a = 100
a = 20 cm
(a) Length of H2 compartment = 80 cm; length of N2 compartment = 20 cm.
(b) Unchanged, remain same]
Q.4(a) Find molality of Ca2+ and NO in 2 M Ca(NO )
aqueous solution of density 1.328 g/mL.
3 3 2
(b) Also find mole fraction of solvent in solution. [3+2]
[Sol: Let volume of solution = 1000ml
W solution = 1000 x 1.328 = 1328 g moles of Ca(NO3)2 = 2
WCa( NO3 )2
= 2 x 164 g = 328 g
mass of water = 1000g
(a) molality of Ca2+ = 328 /164
1
molality of NO – = 2 x124 / 62
= 2 moles
= 4 moles
3 1
55.56
(b) mole fraction of H2O = 6 + 55.56
= 0.901]
Q.5 A 7.38 L vessel containing N2 at 1 atm and 27°C is connected to a 8.2 L vessel containing N2 at 4 atm and 127°C thourgh a stopcock. When stopcock is opened, keeping both vessels at their original temperatures, calculate
(a) Final pressure
(b) Moles of N2 in each vessel. [5]
[Sol: (a) (n1 + n2) initially = (n1 + n2) finally
P1V1 + P2V2 = PV1 + PV2
T1 T2 T1 T2
1x 7.38 + 4 x 8.2
= P 7.38 + 8.2
300
400
300
400
0.0246 + 0.082 = P (0.0246 + 0.0205)
0.1066 = P x 0.0451
P = 2.36 atm
(b) In 1st Vessel
In 2nd vessel
nN2 =
PV1 =
RT1
2.36 x 7.38
0.082 x 300
= 0.708
n = PV2 = 2.36 x 8.2
= 0.59]
N2 RT2
0.082 x 400
Q.6 A common method of working of temperature scales is byassigning temperatures for steam point (Boiling point of water) and absolute zero. A temperature scale commonly used by engineers is the Rankine (R) scale on which absolute zero is 0 R and steam point is 671.4 R. Taking linear variation in temperatur scales.
(a) Calculate ice point (melting point of water) on Rankine scale.
(b) Express 1800 R in Kelvin units. [5]
[Ans: (a) 491.4R (b) 1000K]
Q.7 A container holds 3L of N2(g) & liquid water at 300 K at a pressure of 1 atm. The water in container is
1
instantaneously split into H2 & O2 by electrolysis : H2O(l) H2(g) + 2 O2(g). After the completion
of reaction, the total pressure was 1.698 atm. Calculate mass of water taken initially in the container. Given : Aqueous tension of water at 300 K is 0.04 atm. [6]
[Sol:
PN2 + PH2 = 1atm
PN2 = 1 0.04 = 0.96 atm
H O(l) H (g) + 1 O
(g )
2 2 2 2
P
On completion P 2
after the completion of reaction
P
P + 2 + (0.96 + 0.04) = 1.698
3P 1.396
2 = 0.698; P = 3
1.396 x 3
atm
nH2
formed =
3 x 0.082 x 300
= 0.0567
moles of liquid water = moles of H2 produced = 0.0567
mass of liquid water = 18 x 0.0567 = 1.0206g & for gaseous water
PV = nRT
0.04 x 3 = n x 0.082 x 300
n = 4.878 x 10-3
mass of water vapour = 0.0878
total mass of water taken initially in the container = 1.0206 + 0.0878
= 1.1084 = 1.11gm]
Q.8 Compressbility factor Z vs P plot for four real gases A, B, C & D are shown at 300 K
(a) Out of gases A and B whose molecules are larger? Justify.
(b) If slope of curve for B is 0.02 atm–1, calculate Van der Waals constant b for gas B.
(c) Arrange Boyle temperatures (T) for gases A, B, C & D in increasing order. [2+2+2]
[Sol: (a) Gas A have larger molecular size because it has larger value of Z then B.
(b) Z = 1+ bP
RT
dZ = b
dP RT
= 0.02
b = 0.02 x 0.082 x 300 = 0.492
(c) A < B < C < D]
Q.9 Calculate (in terms of NA)
(a) Number of nitrogen atoms in 160 amu of NH4NO3.
(b) Number of gram-atoms of S in 490 kg H2SO4.
(c) Grams of Al2(SO4)3 containing 3.2 amu of S. [6]
[Sol: (a) NH NO = 160 amu =
160
160
NA gm
2
nNH NO = N
80 = N
4 3 A A
molecules of NH4NO3 = 2
atoms of N = 2 × 2 = 4
(b)
H SO
= 490 Kg = 490 × 103 gm
490 103
nH SO
= 98
= 5000
ns = 5000
3.2 1
(c) ns =
NA 32
= 10NA
1
nAl (SO )
= 30N
2 4 3 A
1 114
= × 342 = gm ]
Al2 (SO4 )3
30NA NA
Q.10 20 g mixture of Na2CO3, NaHCO3 and NaCl on heating to moderate temperatures produced 1.12 L of CO2 at NTP. 8 g of the same mixture in a different experiment required 80 ml of 1 M HNO3 for neutralization. Calculate mass % of each component in mixture. [8]
2NaHCO3
Na2CO3 + H2O + CO2
Na2CO3 + 2HNO3 2NaNO3 + H2O + CO2 NaHCO3 + HNO3 NaNO3 + H2O + CO2
[Sol: Let moles of NaHCO3 in 20 gm mixture = b
2NaHCO3 Na2CO3 + H2O + CO2
nCO
nNaHCO b
= 3 =
2 2
b
2 =
1.12
22.4
= 0.05
b = 0.1
Now in 8 gm mixture
nNaHCO =
b
20 × 8 = 0.4 b
& moles of Na2CO3 = a
Na2CO3 + 2HNO3 2NaNO3 + H2O + CO2
a
& NaHCO3 + HNO3 NaNO3 + H2O + CO2
0.4 b
moles of HNO3 required = 2a + 0.4 b
80 1
1000
= 2a + 0.4 b
2a + 0.4 (0.1) = 0.08
a = 0.02
In 8 gm mixture
Na CO
= 0.02 × 106 = 2.12 gm
2 3
NaHCO
= (0.4 × 0.1) × 84 =3.36 gm
% Na2CO3 =
2.12 100
8
= 26.5%
% NaHCO3 =
3.36 100
8
= 42%
& % NaCl =100 – (26.5 + 42) = 31.5% ]
Q.11 A gaseous mixture containing N2 & H2 in mole ratio 1 : 3 is used to carry out the reaction in a sealed container.
N2 (g) + 3H2(g) 2NH3 (g)
The total pressure was 0.8 atm at 25°C before the reaction. Determine final pressure at 323°C assuming
60% dissociation of reaction. [8]
[Sol: Let
P = P & P
2 2
= 3P (initially)
P + 3P = 0.8 atm P = 0.2 atm
P = 0.2 atm &
2
P = 0.6 atm
2
N2(g) + 3H2(g) 2NH3(g) t = 0 0.2 0.6
't' 0.2(1 – 0.6) 0.6 (1 – 0.6) 2(0.2 × 0.6)
= 0.08 = 0.24 = 0.24
Total pressure = 0.08 + 0.24 + 0.24 = 0.56 atm
Now
P1 = P2
T1 T2
0.56 =
298
P2
596
P2 = 1.12 atm ]
Q.12 Calculate compressibility factor (Z) for 0.02 moles of a Van der Waals gas at pressure of 0.1 atm. Assume the size of gas molecules is negligible.
Given:
RT = 20 L-atm-mol–1 (T = Temperature of gas), a = 1000 atm-L2-mol–2 [8]
n2 a
[Sol:
P +
V 2
(V – nb) = nRT
V – nb ~ V
n2 a
P +
V
n2a
V = nRT
PV + V = nRT
(0.02)2 (1000)
0.1 V +
0.1 V +
0.4
V
V
= 0.4
= 0.02 (20)
0.1 V2 – 0.4 V + 0.4 = 0
V2 – 4V + 4 = 0
(V – 2)2 = 0
V = 2 lt.
Z =
PV nRT =
0.1 2
0.02 20
= 0.5 ]
Q.13 Three ideal gas samples in separate equal volume containers are taken and following data is given:
Pressure Temperature Mean free paths Mol. wt.
Gas A 1 atm 1600 K 0.16 nm 20
Gas B 2 atm 200 K 0.16 nm 40
Gas C 4 atm 400 K 0.04 nm 80
Calculate ratio (A : B : C) of following for the three gases.
(a) Average velocities.
(b) Number of collision made by one molecule per sec. (Z1).
(c) Collision frequencies (Z11). [3+3+3]
[Sol: A =
A2 =
1
=
kTA
kTA
kTB
Similarly 2 =
& C2 =
B
kTC
Now (Vavg)A = =
(Vavg)B = =
(Vavg)C = =
= 200
= 200
= 800
Now NA* =
PA
kTA
PB
1
= 1600k
2 1
*
B kTB
PC
= 200 k
4
= 100k
1
NC* =
kTC
= 400 k
= 100k
2 (V
) N *2
kTA
800 R 1 2
A avg A A
(a) (Z11)A =
kT
200
=
R 1 2
2 1600k
(Z )
= B
11 B
2B PB 100k
kT 200 R 1 2
(Z )
= C
11 C
2C PC
100k
(Z11)A : (Z11)B : (Z11)C
kT 800 R
1 2 kT 200 R 1
2 kT 200 R 1 2
= A
: B
: C
2APA
1600k
2B PB
2 100k
2C PC
2 100k
TA
800
TB
200
TC
200
= P
(1600)2 : P
(100)2 : P (100)2
A A
B B
C C
1600 800 200 200 400 200
= 0.16 1 (1600)2 : 0.16 1 (100)2 : 0.04 4 (100)2
16 8
2 2
4 2
= 0.16 (16)2 : 0.32 : 0.16
= 25 : 25 : 50
8 2
= 1 : 1 : 2
8 2
= 1 : 4 : 16 Ans.
2 *
1
(b) (Z1)A = 2 A (Vavg ) NA =
800
A
1
1600k
(Z1)B =
200
100k
1
(Z1)C =
200
100k
TA
800 TB
200 TC
200
(Z1)A : (Z1)B : (Z1)C = P
:
1600 P
:
100 P
100
A A
B
B
C
C
1600 1 :
200
(2):
400
(2)
= 0.16 0.16 2
0.04 4
1 2
= 50 : 12.5 : 50
= 4 : 1 : 4
(c) (Vavg)A : (Vavg)B :(Vavg)C
= 800
= 4 : 1 : 1
: 200
: 200
SOLUTIONS (CHEMISTRY) RTP-3 ACME
Q.1 Calculate rms velocity of dust particles if mass of one particle of dust 8.31×10–12 g at 329.3°C assuming kinetic theory of gases is applicable. [3]
[Sol: m = 8.31 x 10-12g
or 8.31 x 10-12 x 10-3 = 8.31 x 10-15 kg T = 329.3 x 273 = 6.2.3k
Urms =
= 1.732 x 10-3 m/sec.]
Q.2 7 g of an organic compound on treatment with NaOH gave NH3 which when passed through
1
200 ml of 2 M H2SO4 completely neutralized it. Calculate % of nitrogen in the organic compound.
2NH3 + H2SO4 (NH4)2SO4
[Sol: Org. comp + NaOH NH3 H2 SO4 (NH4)2SO4
[3]
2NH3 + H2SO4 (NH4)2SO4
m. moles of H2SO4 used = 200 x 1/2 = 100
m moles of NH3 produced = 100 x 2 = 200
nNH3 = 0.2
nN = 0.2
wN = 0.2 x 14 = 2.8 gm
% N =
2.8 x100 = 40%]
7
Q.3 A tube of uniform cross-section of length 100 cm is divided into two parts by a weightless and frictionless piston. One part contains 4 moles of hydrogen at 2 atm at equilibrium and other part contains 1 mole of nitrogen at the same temperature. Assume volume of piston to be negligible.
(a) Calculate the length of each compartment if the tube was placed horizontally.
(b) The tube is then held at angle of 45° with the horizontal keeping the nitrogen end upwards. Find the length of each compartment. [1.5+1.5]
[Sol:
Let the area of cross section = A
PV1 = PV2 ;
V1 = V2
= A (100 a) = a A
n1T
n2T
n1 n2 4 1
4a = 100 - a
5a = 100
a = 20 cm
(a) Length of H2 compartment = 80 cm; length of N2 compartment = 20 cm.
(b) Unchanged, remain same]
Q.4(a) Find molality of Ca2+ and NO in 2 M Ca(NO )
aqueous solution of density 1.328 g/mL.
3 3 2
(b) Also find mole fraction of solvent in solution. [3+2]
[Sol: Let volume of solution = 1000ml
W solution = 1000 x 1.328 = 1328 g moles of Ca(NO3)2 = 2
WCa( NO3 )2
= 2 x 164 g = 328 g
mass of water = 1000g
(a) molality of Ca2+ = 328 /164
1
molality of NO – = 2 x124 / 62
= 2 moles
= 4 moles
3 1
55.56
(b) mole fraction of H2O = 6 + 55.56
= 0.901]
Q.5 A 7.38 L vessel containing N2 at 1 atm and 27°C is connected to a 8.2 L vessel containing N2 at 4 atm and 127°C thourgh a stopcock. When stopcock is opened, keeping both vessels at their original temperatures, calculate
(a) Final pressure
(b) Moles of N2 in each vessel. [5]
[Sol: (a) (n1 + n2) initially = (n1 + n2) finally
P1V1 + P2V2 = PV1 + PV2
T1 T2 T1 T2
1x 7.38 + 4 x 8.2
= P 7.38 + 8.2
300
400
300
400
0.0246 + 0.082 = P (0.0246 + 0.0205)
0.1066 = P x 0.0451
P = 2.36 atm
(b) In 1st Vessel
In 2nd vessel
nN2 =
PV1 =
RT1
2.36 x 7.38
0.082 x 300
= 0.708
n = PV2 = 2.36 x 8.2
= 0.59]
N2 RT2
0.082 x 400
Q.6 A common method of working of temperature scales is byassigning temperatures for steam point (Boiling point of water) and absolute zero. A temperature scale commonly used by engineers is the Rankine (R) scale on which absolute zero is 0 R and steam point is 671.4 R. Taking linear variation in temperatur scales.
(a) Calculate ice point (melting point of water) on Rankine scale.
(b) Express 1800 R in Kelvin units. [5]
[Ans: (a) 491.4R (b) 1000K]
Q.7 A container holds 3L of N2(g) & liquid water at 300 K at a pressure of 1 atm. The water in container is
1
instantaneously split into H2 & O2 by electrolysis : H2O(l) H2(g) + 2 O2(g). After the completion
of reaction, the total pressure was 1.698 atm. Calculate mass of water taken initially in the container. Given : Aqueous tension of water at 300 K is 0.04 atm. [6]
[Sol:
PN2 + PH2 = 1atm
PN2 = 1 0.04 = 0.96 atm
H O(l) H (g) + 1 O
(g )
2 2 2 2
P
On completion P 2
after the completion of reaction
P
P + 2 + (0.96 + 0.04) = 1.698
3P 1.396
2 = 0.698; P = 3
1.396 x 3
atm
nH2
formed =
3 x 0.082 x 300
= 0.0567
moles of liquid water = moles of H2 produced = 0.0567
mass of liquid water = 18 x 0.0567 = 1.0206g & for gaseous water
PV = nRT
0.04 x 3 = n x 0.082 x 300
n = 4.878 x 10-3
mass of water vapour = 0.0878
total mass of water taken initially in the container = 1.0206 + 0.0878
= 1.1084 = 1.11gm]
Q.8 Compressbility factor Z vs P plot for four real gases A, B, C & D are shown at 300 K
(a) Out of gases A and B whose molecules are larger? Justify.
(b) If slope of curve for B is 0.02 atm–1, calculate Van der Waals constant b for gas B.
(c) Arrange Boyle temperatures (T) for gases A, B, C & D in increasing order. [2+2+2]
[Sol: (a) Gas A have larger molecular size because it has larger value of Z then B.
(b) Z = 1+ bP
RT
dZ = b
dP RT
= 0.02
b = 0.02 x 0.082 x 300 = 0.492
(c) A < B < C < D]
Q.9 Calculate (in terms of NA)
(a) Number of nitrogen atoms in 160 amu of NH4NO3.
(b) Number of gram-atoms of S in 490 kg H2SO4.
(c) Grams of Al2(SO4)3 containing 3.2 amu of S. [6]
[Sol: (a) NH NO = 160 amu =
160
160
NA gm
2
nNH NO = N
80 = N
4 3 A A
molecules of NH4NO3 = 2
atoms of N = 2 × 2 = 4
(b)
H SO
= 490 Kg = 490 × 103 gm
490 103
nH SO
= 98
= 5000
ns = 5000
3.2 1
(c) ns =
NA 32
= 10NA
1
nAl (SO )
= 30N
2 4 3 A
1 114
= × 342 = gm ]
Al2 (SO4 )3
30NA NA
Q.10 20 g mixture of Na2CO3, NaHCO3 and NaCl on heating to moderate temperatures produced 1.12 L of CO2 at NTP. 8 g of the same mixture in a different experiment required 80 ml of 1 M HNO3 for neutralization. Calculate mass % of each component in mixture. [8]
2NaHCO3
Na2CO3 + H2O + CO2
Na2CO3 + 2HNO3 2NaNO3 + H2O + CO2 NaHCO3 + HNO3 NaNO3 + H2O + CO2
[Sol: Let moles of NaHCO3 in 20 gm mixture = b
2NaHCO3 Na2CO3 + H2O + CO2
nCO
nNaHCO b
= 3 =
2 2
b
2 =
1.12
22.4
= 0.05
b = 0.1
Now in 8 gm mixture
nNaHCO =
b
20 × 8 = 0.4 b
& moles of Na2CO3 = a
Na2CO3 + 2HNO3 2NaNO3 + H2O + CO2
a
& NaHCO3 + HNO3 NaNO3 + H2O + CO2
0.4 b
moles of HNO3 required = 2a + 0.4 b
80 1
1000
= 2a + 0.4 b
2a + 0.4 (0.1) = 0.08
a = 0.02
In 8 gm mixture
Na CO
= 0.02 × 106 = 2.12 gm
2 3
NaHCO
= (0.4 × 0.1) × 84 =3.36 gm
% Na2CO3 =
2.12 100
8
= 26.5%
% NaHCO3 =
3.36 100
8
= 42%
& % NaCl =100 – (26.5 + 42) = 31.5% ]
Q.11 A gaseous mixture containing N2 & H2 in mole ratio 1 : 3 is used to carry out the reaction in a sealed container.
N2 (g) + 3H2(g) 2NH3 (g)
The total pressure was 0.8 atm at 25°C before the reaction. Determine final pressure at 323°C assuming
60% dissociation of reaction. [8]
[Sol: Let
P = P & P
2 2
= 3P (initially)
P + 3P = 0.8 atm P = 0.2 atm
P = 0.2 atm &
2
P = 0.6 atm
2
N2(g) + 3H2(g) 2NH3(g) t = 0 0.2 0.6
't' 0.2(1 – 0.6) 0.6 (1 – 0.6) 2(0.2 × 0.6)
= 0.08 = 0.24 = 0.24
Total pressure = 0.08 + 0.24 + 0.24 = 0.56 atm
Now
P1 = P2
T1 T2
0.56 =
298
P2
596
P2 = 1.12 atm ]
Q.12 Calculate compressibility factor (Z) for 0.02 moles of a Van der Waals gas at pressure of 0.1 atm. Assume the size of gas molecules is negligible.
Given:
RT = 20 L-atm-mol–1 (T = Temperature of gas), a = 1000 atm-L2-mol–2 [8]
n2 a
[Sol:
P +
V 2
(V – nb) = nRT
V – nb ~ V
n2 a
P +
V
n2a
V = nRT
PV + V = nRT
(0.02)2 (1000)
0.1 V +
0.1 V +
0.4
V
V
= 0.4
= 0.02 (20)
0.1 V2 – 0.4 V + 0.4 = 0
V2 – 4V + 4 = 0
(V – 2)2 = 0
V = 2 lt.
Z =
PV nRT =
0.1 2
0.02 20
= 0.5 ]
Q.13 Three ideal gas samples in separate equal volume containers are taken and following data is given:
Pressure Temperature Mean free paths Mol. wt.
Gas A 1 atm 1600 K 0.16 nm 20
Gas B 2 atm 200 K 0.16 nm 40
Gas C 4 atm 400 K 0.04 nm 80
Calculate ratio (A : B : C) of following for the three gases.
(a) Average velocities.
(b) Number of collision made by one molecule per sec. (Z1).
(c) Collision frequencies (Z11). [3+3+3]
[Sol: A =
A2 =
1
=
kTA
kTA
kTB
Similarly 2 =
& C2 =
B
kTC
Now (Vavg)A = =
(Vavg)B = =
(Vavg)C = =
= 200
= 200
= 800
Now NA* =
PA
kTA
PB
1
= 1600k
2 1
*
B kTB
PC
= 200 k
4
= 100k
1
NC* =
kTC
= 400 k
= 100k
2 (V
) N *2
kTA
800 R 1 2
A avg A A
(a) (Z11)A =
kT
200
=
R 1 2
2 1600k
(Z )
= B
11 B
2B PB 100k
kT 200 R 1 2
(Z )
= C
11 C
2C PC
100k
(Z11)A : (Z11)B : (Z11)C
kT 800 R
1 2 kT 200 R 1
2 kT 200 R 1 2
= A
: B
: C
2APA
1600k
2B PB
2 100k
2C PC
2 100k
TA
800
TB
200
TC
200
= P
(1600)2 : P
(100)2 : P (100)2
A A
B B
C C
1600 800 200 200 400 200
= 0.16 1 (1600)2 : 0.16 1 (100)2 : 0.04 4 (100)2
16 8
2 2
4 2
= 0.16 (16)2 : 0.32 : 0.16
= 25 : 25 : 50
8 2
= 1 : 1 : 2
8 2
= 1 : 4 : 16 Ans.
2 *
1
(b) (Z1)A = 2 A (Vavg ) NA =
800
A
1
1600k
(Z1)B =
200
100k
1
(Z1)C =
200
100k
TA
800 TB
200 TC
200
(Z1)A : (Z1)B : (Z1)C = P
:
1600 P
:
100 P
100
A A
B
B
C
C
1600 1 :
200
(2):
400
(2)
= 0.16 0.16 2
0.04 4
1 2
= 50 : 12.5 : 50
= 4 : 1 : 4
(c) (Vavg)A : (Vavg)B :(Vavg)C
= 800
= 4 : 1 : 1
: 200
: 200
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