Chemical Bonding

GASEOUS STATE

 

Hints to Subjective Problems

 

LEVEL – I

1. a) PV =
2. b) Apply
3. Apply r ∝
4. a) Calculate the mass of an electron.
5. Calculate empirical formula first.
6. 2nd contraction is because of O2.
7. a) Calculate total O2 required for the process.

 

LEVEL – II

1. a) no. of moles are same initially and finally.
2. b) Calculate the volume of H2
3. a) Calculate the density of mixture first.
4. b) Follow curve
5. Calculate the no. of moles of all gases at the given conditions.
6. b) Water gas is CO + H2
7. Calculate density of mix and then apply

LEVEL – III

1. Apply
2. Calculate the no. of moles of N2 and I2 is vessel.
3. No. of moles at both the ends is same.
4. Pressure in all the balloons is same.
5. Calculate the no. of atoms from the last breath of plato per m3.
6. Calculate meff. of ozonised air.
7. Keep Lung capacity same
8. Calculate total no. of moles of A and A2.
9. Apply Dalton’s Law.
10. Differentiate the equation E =

Solution to Subjective Problems

 

LEVEL – I

1. a) PV =

⇒ V = 25.565 litre

1. b) PM = dRT

⇒ d = 1.5124 g/lt

2. a) Mol. wt. (M)  = ×22.4; V = Vol. In lit. apply pV = RT

w = Wt. of the gas in g at 473 K and 16 atm]

 

1. b) According to gas equation,

P2 = = 1051 mm

3. a) H2 > He > N2

Rate of diffusion of the gas and

b)

∴

ratio of mole fraction = = =

∴1: 24

4. a) Mass of electron on atomic weight scale is amu

Mass of 1 mole of electron =

We have C = =

∴ r.m.s. speed of an electron = 1.16 × 107 cm/sec

To calculate rms speed of UF6 put M = 238 + 6 × 19 = 352

b)

= 1.632

5 a) K.E. of one mole of a gas = = = 900 calories  

1. b) Change in translation  kinetic energy per mole

= = = 300 cals.

Change in translation  kinetic energy for 3.45 g Ne

= = 51.29 cal.

6. Vc = 3b = and also

Vc =

So

⇒ r3 = 4.25 × 10-30

⇒ r = 1.62 Ao

7. b =

r = = 1.47 × 10–8 cm

8. On combustion in O2 all carbon and hydrogen atoms of the compound are converted into CO2 and H2O respectively.

44 gm CO2 having 12 gm C

∴ 0.195 gm CO2 having 0.0531 gm C

Similarly 0.0804 gm H2O contain 0.0089 gm H2.

Hence 0.22 gm compound contain 0.0531 gm C, 0.0089 gm H2 & 0.158 gm H2

∴ % of C = 24.13%; H = 4.04% Cl = 71.8%

or V weights = 12 gm

22400 ml weights 98.5 gm

The empirical formula is CH2Cl

Molecular formula (CH2Cl)n

Let volume of 0.12g compound occupies V ml at N.T.P. then

Molecular weight = 98.5

∴ 12 n + 2n + 35.5 n = 49.5n = 98 or n = 2

∴ molecular formula = C2H4Cl2

9. Reactions are CH4 + 2O2 ⎯→ CO2 + 2H2O

C2H4 + 3O2 ⎯→ 2CO2 + 2H2O

Let volume of CH4, C2H4 x and y ml

∴ volume of CO2 = 10 – (x + y)ml

1st volume contraction = (x + 2y + y + 3y) – (x + 2y) = 2x + 2y

By question, 2x + 2y = 17

or 2x = 17 – 8 = 9 or x = 4.5

Because y = 4 [total volume of CO2 = x + 2y + 10 – x – y = 14]

∴ y = 4

Hence volume of C2H4 = 4.5 ml

Volume of CH4 = 4 ml

Volume of CO2 = 1.5 ml

% of CH4 = = 45%

% of C2H4 = = 40%

% of CO2  = = 15%

10. a) Let the volume of gas mixture at N.T.P. be V ml

V ml = 2730 ml

According to problem

100 ml mixture contain 20 ml methane 60 ml CO and 20 ml H2

∴ Volume of methane in 2730 ml of mixture = = 546 ml

Volume of CO= = 1638 ml

Volume of Hg = (2730 – 546 – 1638) = 546 ml

The reactions are

CH4 + 2O2 ⎯→ CO2 + H2O …(1)

2CO + O2 ⎯→ 2CO2 …(2)

2H2 + O2 ⎯→ 2H2O …(3)

∴ Total volume of O2 required for oxidation of

546 ml CH4, 1638 ml CO & 546 ml H2 are

= ml = 2184 ml

Again 2KClO3 ⎯→ 2KCl + 3O2

3 × 22.4 lit of O2  at NTP are obtained from 245 gm of KClO3

2.184 lit of O2 at NTP are obtained from of KClO3 = 7.96g of KClO3

1. b) When the mixture of gases is passed over hot copper, N2 will remain unchanged but NO will be reduced by Cu to N2

2Cu + 2NO ⎯→ 2CuO + N2

Let x cc of N2 be present in 25cc mixture. Hence the volume of NO in the mixture is (25 – x)cc

From the equation under same conditions of temperature and pressure 2 volume of NO gives 1 volume of N2

(25 – x)cc NO will give cc of N2

Total volume after the reaction = cc

By question, = 20 or x = 15cc

So the mixture contains 15cc N2 and (25 – 15) or 10cc NO

Hence percentage of N2 = = 60%

Percentage of NO = = 40% 

 

LEVEL – II

1. a) Let the volume of each bulb be V lit

For two connected bulbs P = 0.5 atm. n = 0.7 mole

T = 300 K. Volume = 2V 

We have PV = nRT

0.5 (2V) = 0.7 × 0.082 × 300

or V = 17.22 lit

When one of the bulbs is maintained at 127°C i.e. 400K & other at 300 K let the moles of H2 in these bulbs be n1 & n2 respectively.

n1 + n2 = 0.7 …(1)

Since stopcock is open, the pressure in each bulb

Will be same let it be p atm. Thus for bulb at 400K

pV = n1 × 0.082 × 400 

or  17.22 p = 32.8 n1 or p = 1.9 n1 …(2)

For 2nd bulb at 300 K

p × 17.22 = n1 × 0.082 × 300

p = 1.42 n2 …(3)

From equation (1), (2) and (3) p = 0.571 atm

n1 = 0.3 mole n2 = 0.4 mole

1. b) Mg + 2HCl ⎯→ MgCl2 + H2

24g 22.4 lit at N.T.P.

2Al + 6HCl ⎯→ 2AlCl3 + 3H2

2 × 27 = 54g 3 × 22.4 lit at N.T.P.

Let V be volume of liberated H2 at N.T.P.

∴ or V = 1103.7 ml

Let weight of Mg in alloy be x gm ∴ wt. of Al = (1 – x)g

∴ By problem, (1 – x) = 1103.7

∴ x = 0.4525, % of Al = 54.75%

∴ Wt. of Mg = 0.4525g

Wt. of Al = (1 – 0.4525)g = 0.5475gm

% of Mg = 45.25

2. a) Let V be the volume diffusing out in each case

Let d0.dm and d be the densities of the mixture of pure oxygen and ozone respectively.

∴

∴ dm = 16 ≈ 21.91

Again for 100 gms the volume =

or ∴ d = 23.3

1. b) From the extrapolation of vs P curve to P = 0 the value of is found to be 1.517 gm lit–1.

∴ = 1.517 × 0.082 × 273 = 34

∴ The atomic weight of P = 34 – 3 × 1.008 = 30.98

3. i) Volume of 1 molecule = πr3

(Q r = 150 pm = 150 × 10–10 cm)

Volume of 1 molecule =

V1 = 1.41 × 10–23 cc/molecule

1. ii) The volume occupied by N molecules = N × V1

= 6.023 × 1023 × 1.41 × 10–23

= 8.49 cc per mole

∴ Volume of 1 mole of N2 = 22400 ml of STP

Thus empty space = (22400 – 8.49) = 22391.51cc

% empty space = × 100 = 99.96%

4. i) CxHyNz + O2 ⎯→ CO2+ H2O (Vapour) + N2

(suppose the compound is CxHyNz)

∴ (9 – V) volume V volume 4 volume 6 volume 2 volumes

(9 – V) mole V moles 4 moles 6 moles 2 moles

By 2V = 2 × 4 + 1× 6 = 14 or, V = 7 volume

1. ii) CxHyNz + O2 ⎯→ CO2 + H2O (vapour) N2

2 moles 7 moles 4 moles 6 moles 2 moles

∴x × moles of CxHyNz = V moles of CO2

x + 2 = 1 × 4 x = 2

xyz moles of CxHyNz = 2 × moles of H2O (vapour)

x moles of CxHyNz = 2 × moles of N2

z × 2 = 2 × 2 , z = 2

Here the compound is C2H6N2

5. Applying PV = nRT

for NO, × 0.25 = n1RT ∴n1 =

for O2, = n2RT n2 =

Since NO & O2 react in 2 : 1 molar ratio

2NO + O2 ⎯→ N2O4

Here O2 is the limiting reagent

Mole of NO remain after completion of reaction

=

The pressure due to remaining NO is

p = or p = 0.30 atm

Wt. of N2O4 = mole of N2O4 × 92

= = 0.402g

6. a) i) for ideal gas nature Z = 1

PV = nRT

V = =

V = 0.0448 litre

1. ii) for real gas

PV = ZnRT

V = =

V = 8.987 × 10-3 litre.

b)

Dividing by (V-nb) and solving for P,

Substituting n = 1, R = 0.0821 L atm mol-1 K-1, V = 0.25 L, T = 300 K and the values of a and b, we have

 

From the ideal gas equation,

P =  = 98.52 atm

7. a) Vescape = = O2

=  use R = 8:314

T = = = 1.6 ×105K

 

36. b) Molecular mass of HCl = 36.5

Molecular mass of NH3 = 17.0

According to Graham’s law of diffusion,

Let the distance between P and X be d cm.

∴ = 0.6824

or d = (0.6824) (200 – d)

or d + 0.6824 d = 0.6824 × 200

or d = = 81.12 cm

8. a) Let weight of Cx H8 = y gm ∴ weight of CxH12 = (41.4–y) g 

1.56 ×10 = ×0.082 ×317 ——— (1)

% of hydrogen  = = 14 ——— (2)

On solving equation (I) and (2) we get 

x =  5 and  y = 23.8

(1) Gases are C5H8 and C5H12

(2) C5H8 = 0.35 mole C5H12 = 0.244 mole 

1. b) Combustion equations :

(i) 2H2 + O2 → 2H2O

2 vols. + 1 vol Zero (liquid)

(ii) 2CO + O2 → 2CO2

2 vols. 1 vol. 2 vols.

 

From the equations, it is clear that 12 ml. of H2 present in 24 ml. of water gas will react with 6 ml. of oxygen to produce negligible volume of liquid water at room temperature and 12 ml. of CO present in the water gas will react with 6 ml. of O2 to produce 12 ml. of CO2.

Thus, the total volume of oxygen used = 6 + 6 = 12 ml.

Volume of O2 present in 80 ml. of air containing 20% oxygen =

Hence, the volume of gases at the end of the reaction will be:

Oxygen unused = 16 −12 = 4ml.

Nitrogen unused = 80 − 16 = 64 ml.

and carbon dioxide produced = 12 ml.

[NOTE: Volume of steam produced in equation(i) above will be the same as the volume of H2 used, but when it is condensed at room temperature to water, the volume occupied by it is negligible.]

9. At 273 + 27 = 300K (T1), let the number of moles of air be 1 and at T2 the number of moles would be 1 – 3/5 = 2/5 = 0.4 moles. Pressure is constant, because the vessel is open and the volume of the vessel is also constant (given).

Using the gas equation: PV = n1RT1 = n2RT2

Whence, = T1/T2 = n2/n1 or T2 = T1 × n1/n2 = = 750K

10. N2O4 2NO2

1 – α 2α Total no. of moles = 1 + α

The calculated density of N2O4  = 46

Observed density ρ  = 30.2

Hence α = = = 0.523

The fraction of gas molecules decomposed = 0.523

% of NO2 molecules by weight = 52.3%

Again in the mixture the ratio of gm moles of NO2 and N2O4 is

∴ % of NO2 by volume = × 100 = 68.7%

 

LEVEL – III

 

1. Applying PV = nRT =

or P = where S = solubility

∴ = constant for a gas at a given temperature

∴

Given that P1 = 760 mm, P2 = 740 mm

S2 = = 0.97S1

For oxygen solubility at 740 mm = 0.97 × 0.048 = 0.0465 g ml–1

= 0.0465 g ml–1

For nitrogen solubility at 740 mm = 0.97 × (0.02)2 = 0.0213 g ml–1

For CO solubility at 740 mm = 0.97 × 0.03 = 0.029 g ml–1

∴ Amount of O2 dissolved in 200 ml of water = (0.0465 × 0.2) = 0.0093 g ml–1

Amount of N2 dissolved in 200 ml of water = 0.0213 × 0.2 = 0.00426 g ml–1

Amount of CO dissolved in 200 ml of water = 0.029 × 0.2 = 0.0058 g ml–1

2. 10 gms of I2 (solid) and Nitrogen at 10 atm pressure and at 25°C is taken in one litre flask. The pressure is due to only Nitrogen as I2 is solid at 25° C

Applying gas equation

PV = nRT, we determine the moles of Nitrogen taken

P = 10 atm V = 1 litre R = 0.0821 T = 25 + 273 = 298 K

10 × 1 = × 0.0821 × 298

= 0.4087 moles

Moles of I2(s) taken = = 0.0394

Total moles of I2(s) and Nitrogen in I flask of volume 1 litre is,

= 0.4087 +   0.0394 =  0.4481 moles

On heating the I flask to 523 K I2(s) vaporises and we get a mixture of I2 (vapours) and Nitrogen. Out of 0.4481 moles, let n moles migrate to second flask kept at 473 K. Let Pnew be the pressure in the flask developed.

Applying gas equation

Pnew × 1 = (0.4481 –n) × R × 523 …(1)

Pnew × 40 = n × R × 473 …(2)

Dividing (1) by (2) we get,

On solving we get,

n = 0.4382 moles

Substituting the value of ‘n’ in equation (2) we get,

Pnew × 40 = 0.4382 × 0.0821 × 473

Pnew = 0.4254 atm

3. Let initially the length of air column on each side be L, then 2L + 10 = 100

∴ L = 45 cm

If the tube is held vertifically, let the Hg column be displaced downwards by y to attain same pressure above and below the column of Hg. Then

PB + 10 = PA (Pressure an taken in terms of length of Hg) …1

For end A: Since mole remains same on two sides

…(2) (V = L × area)

For end B: = …(3)

From (1), (2) and (3) L = 45 cm, P0 = 76 cm y = 3 cm

4. Let initially the length of air column in tube be L cm, then

2L + 5 = 46 + 5 + 44.5

∴ L = 45.25 cm

When the tube is held vertically at 60.4 the Hg will be displaced to lower end. So that 

PB + 5cos60° = PA

or PA – PB= 2.5 cm …(1)

For end A: …(2)

For end B: …(3)

From equation (1), (2) and (3)

PO = 75.4 cm of Hg

5. 5. Convert the observed pressure drop to mm Hg:   (37 mm) mm Hg in 30 min, or 5.6 mmHg drop per hour. With 5.6 mm Hg as the oxygen partial pressure, change to STP:
6. Volume of flask A =

Volume of balloon B =

Volume occupied by H2 in A =

= 3.284 m3 = 3284 lit

Pressure in A = = 1.123 atm

Pressure in B = 1.123 atm

Volume of N2  in B =

= 0.7917 m3 = 791.7 lit

Moles of N2 = = 36.14

Mass of N2 = (36.14 × 28) = 1011.92g

Mass of O2 in C = = 165.12g

Total weight of all the gases (165.2 + 300 + 1011.92)g = 1477.12g

Total pressure inside the flask of –78°C after the balloon cracks up

= = 0.73 atm

7. n = = 0.0196 mole air. breath–1

Number of atoms of Ar(N) = 0.01 × 0.0196 × 6.023 × 1023 

= 1.18 × 1020 breath–1

The concn of atoms from the last breath of Plato is = 23.6 m–3

Giving the number of breaths needed as  = 84.7 breaths 

which is equivalent to  = 8.47 min

8. For CO2 output rate = 44 g per hour

= 1 mole/hr

= 22.4 lit/hr at STP

CO2 reduction rate = 600 ml/min

600 × 60 ml/hr = 36 lit/hr

Fraction of time by which converter has to be operated =

= = 0.622

9.

M = 30.3

When 100 ml of air became 95 ml ozonized air 

3O2   2O3

21 – 3x         2x 

volume reduction , x = 5 ml

O2 left = 21 – 15 = 6 ml

O3 formed , 2x = 10

30.3 =

M1 : Mol. wt. of ozone

Mol. wt. of O3  = 48.26 

10. Since Ammonia on sparking will decompose to form Nitrogen and Hydrogen only, the volume of 20 ml. after sparking consists of a mixture of Nitrogen and Hydrogen. Let the volume of Hydrogen be x ml., therefore, the volume of Nitrogen will be (20 − x) ml. When mixed with 30 ml. of oxygen and exploded, the hydrogen of the mixture reacted with oxygen and a contraction of (30 + 20−27.5) = 22.5 ml. took place. According to the equation.

  2H2 + O2 → 2H2O

2 vols. 1 vol. zero vol. (after cooling)

or x ml. x/2 ml.

Thus, contraction in volume   =  

    ∴ x = 15

Thus, 10 ml. of ammonia, on decomposition, produced 15 ml. of hydrogen and 20−15=5 ml of Nitrogen. or 2 vols. of Ammonia on decomposition, produced 3 vols. of Hydrogen and 1 vol. of Nitrogen.

By Avogadro’s hypothesis:

2 molecules of ammonia on decomposition produce 3 molecules or 6 atoms of hydrogen and 1 molecule or 2 atoms or nitrogen.

Thus the formula of ammonia should be NH3.

V.D. of ammonia = 8.5

∴ Mol. wt. of ammonia = 8.5 × 2 = 17

Mol. wt. is also in accordance with the formula of ammonia – NH3 

11. Molecules of O2 required per hour by the body = 3.24 × 1022

∴ moles of O2 required per hour = = 0.0538

Moles of O2 inhaled in five minutes =

Moles of air taken in five minutes =

∴ Lung capacity at the himalayan cave  =

Since the lung capacity remains the same,  therefore the number of moles of air taken in one breathing  

=

Therefore, moles of O2 taken in one breathing =

If the number of breathing in one hour is x

then 4.86 × 10-3 x = 0.0538 

x = 11.067

Breathing rate = = 5.42 per min per breath.

12. At 273°C, the wt. of A= = 48 g = 1  mole (being 50%)

Also at 273°C, the wt of A = 48 g = moles = 0.5 mole 

(since Mol. wt. of  A2 = 48 × 2 = 96)

Total No. of moles at 273° C = 1 +0.5 = 1.5 moles.

By gas equation PV = nRT, where n = No. of moles or

= 2atm 

13. Volume of container = 0.731 ml

Temperature = 23 + 273 = 296 K

Initial pressure = = 1.74 mm …(1)

At  –750C, H2O is frozen out and the mixture on retaining 23°C has pressure P

P = = 1.32 mm …(2)

At –95°C, CO2 is also frozen out and the mixture on retaining 23°C has pressure

= 0.53 mm …(3)

By (2) and (3) = 1.32  0.53 = 0.79 mm

By (1) and (2) = 1.74  1.32 = 0.42 mm

Now using PV = nRT for each gas separately

For N2: = 2.1 × 10-8

For CO2: n = = 3.1× 10-8

For (H2O) v n =

= = 1.7 × 10–8 

14. The number of molecules per cc at  25°C & 1 atm 

n = = 2.46 × 1019

The average velocity CAV = = 4.8 ×104 cm/sec

Hence no. of binary collisions per cc per sec

ZAA =

= × (3.74 × 10–8)2 × 4.8 ×104 × (2.462 × 1019)2

= 8.98 × 1028

15. We have 

= 6.2 × 10–14 ergs

dE = 0.01 E = 6.2 × 10–16 ergs

= × e–3/2 × (6.2 × 10–16)

= 0.0046 ∴ 46%

 

Solution to Objective Problems

 

LEVEL – I

1.

⇒ V2 = 4V1

∴ (D)

2. = 4

∴ (A)

4. or

or

∴ (B)

7. Gases with the same molecular weight diffuses at the same rate 

∴ CO2 and N2O (molecular weight 44)

∴ (A)

9. = = 8:1

∴ (C)

11. = nRT or PV = nRT =

∴ (A)

13. Differentiate Charle’s Equation 

∴ (D)

14.

∴ (D)

LEVEL – II

2. More is the value of a mass is the liquifaction and bigger is the size of molecule bigger is the ‘b’  value

∴ (B)

3. CxHy + O2 ⎯→ xCO2 + H2O

0.2 

0                     0.2x      × 0.2

given 0.2x = 8

x = 4

0.1y = 0.1

y = 10

∴ C4H10

∴ (A)

4. After the introduction of 3g of B the pressure was became half it means no. of moles of B 

or

or

∴ (C)

5. Under similar conditions

n1T1 = n2T2

n1 × 600 = n1 × T2

T2 = 750K

T2 = 477°C

∴ (D)

6. PV =

M = 256

∴atomicity = = 8

∴ (D)

7.

T2 = 372.5 K

= 99.5°C

∴ (C)

8.

2  =

T2= 273 × 4 = 1092 K

∴ (D)

11. K.E. ∝ T

∴ (B)

12. Calculate Meff.

∴ (A)

14. Vrms ∝

∴ (D)

15. where Ax & Ay are the area of cross section of orifice.

∴ r = radius of a radius orifice

= length of square orifice

∴ (A)

 

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