Radius of the balloon = 10 m
\(\therefore\)Volume of the balloon = \(\frac{4}{3}\pi\)r² = \(\frac{4}{3}\times\frac{22}{7}\times\)(10m)³ = 4190.5 m³
Volume of He filled at 1.66 bar and 27°C = 4190.5 m3
Calculation of mass of He
PV = nRT = \(\frac { w }{ M } \)RT
or \(w\) = \(\frac{MPV}{RT}\) = \(\frac { (4\times { 10 }^{ -3 }kgmol^{ -1 })(1.66bar)(4190.5\times 103dm^{ 3 }) }{ (0.083bardm^{ 3 }K^{ -1 }mol^{ -1 })(300K) } \)
= 1117.5 kg
Total mass of the balloon along with He = 100 + 1117.5 = 1217.5 kg
Maximum mass of the air that can be displaced by balloon to go up = Volume x Density
= 4190.5 m³ x 1.2 kg m^-3 = 5028.6 kg
\(\therefore\) Pay load = 5028.6 - 1217.5 kg = 3811.1 kg

(i) Certain plants, e.g. Pinus, Juniparus, Quercus, Pyrus and Vitis can metabolise nitrogen oxide (NO) and therefore, their plantation could help in reducing photochemical smog.
(ii) Liquefied CO₂ with a suitable detergent is used for dry cleaning and H₂O₂ is used for the better results and makes use of lesser amount of water.
(iii) Soaps are 100% biodegradable so they should be used in place of detergents. Now-a-days biodegradable detergents are available. Therefore, they should be used in place of non-biodegradable hard detergents.
(iv) CNG should be used as it causes much less pollution. Moreover, electrical vehicles should be used to reduce the consumption of petrol and diesel

Since, a colourless liquid A contains only H and O and decomposes slowly on exposure to light but is stabilised by additions of urea, therefore, liquid A may be hydrogen peroxide.
(ii) \(2{ H }_{ 2 }{ O }_{ 2 }\left( l \right) \underrightarrow { hv } 2{ H }_{ 2 }O\left( l \right) +{ O }_{ 2 }\left( g \right) \)

(a) Writing the On on each atom above its symbol, then
\(3\overset { +1 }{ H } \overset { -1 }{ Cl } (aq)+\overset { +1 }{ H } \overset { +5 }{ N } \overset { -2 }{ { O }_{ 3 } } (aq)\longrightarrow \overset { 0 }{ { Cl }_{ 2 }(g)+ } \overset { +3 }{ N } \overset { -2 }{ O } \overset { -1 }{ Cl(g)+ } \overset { +1 }{ { 2H }_{ 2 } } \overset { -2 }{ O } (l)\)
Here, the On of Cl increases from -1 in HCl to O in Cl₂ , therefore, Cl^- is oxidised and hence, HCl acts as the reducing agent. The ON of N decreases from +5 in HNO₃ to +3 in NOCL, therefore, HNO₃ acts as the oxidising agent. Thus this reaction is a redox reaction.
(b) Writing the ON of each atom above its symbol, we have,
\(\overset { +2 }{ Hg } \overset { -1 }{ { Cl }_{ 2 } } (aq)+\overset { +1 }{ 2K } \overset { -1 }{ I } (aq)\longrightarrow \overset { +2 }{ Hg } { \overset { -1 }{ I } }_{ 2 }(s)+2\overset { +1 }{ K } \overset { -1 }{ { Cl }^{ - }(aq) } \)
Here, the On of none of the atoms undergo a change, therefore, this reaction is not a redox reaction.
(c) \(\overset { +3 }{ { Fe }_{ 2 } } { \overset { -2 }{ O } }_{ 3 }(s)+3\overset { +2 }{ C } \overset { -2 }{ O } (g)\ \overset { \Delta }{ \longrightarrow } \ 2\overset { 0 }{ Fe } (s)+3\overset { +4 }{ C } \overset { -2 }{ { O }_{ 2 } } (g)\)
Here, On of decreases from +3 in Fe₂O₃ to 0 in Fe, therefore, Fe₂O₃ acts as an oxidising agent. Further, On of C increases from +2 in CO to +4 in CO₂, therefore, CO acts as a reducing agent. Thus, this reaction is an example of redox reaction.
(d) Writing the ON of each atom above its symbol, then
\(\overset { +3 }{ P } { \overset { -1 }{ Cl } }_{ 3 }(l)+3\overset { +1 }{ { H }_{ 2 } } \overset { -2 }{ O } (l)\quad \longrightarrow \quad 3\overset { +1 }{ H } \overset { -1 }{ Cl } (aq)+\overset { +1 }{ { H }_{ 3 } } \overset { +3 }{ P } \overset { -2 }{ { O }_{ 3 } } (aq)\)
Here, On of none of the atoms undergo a change, therefore, this reaction is not a redox reaction.
(e) Writing the ON of each atom above its symbol, then
\(4\overset { -3 }{ N } \overset { +1 }{ { H }_{ 3 } } (aq)+3\overset { 0 }{ { O }_{ 2 } } (g)\quad \longrightarrow \quad 2\overset { 0 }{ { N }_{ 2 } } (g)+6\overset { +1 }{ { H }_{ 2 } } \overset { -2 }{ O } (l)\)

We know that, \({ K }_{ p }={ K }_{ c }{ \left( RT \right) }^{ \triangle ng }\)
For the above reaction,
\(\triangle ng=\left( 2+1 \right) -2=1\)
\({ K }_{ p }=3.75\times { 10 }^{ -6 }\left( 0.0831\times 1069 \right) \)
K_p = 0.033

For the reaction, \({ PCl }_{ 5 }(g)\rightleftharpoons { PCl }_{ 3 }(g)+{ Cl }_{ 2 }(g)\)
At 500 K in a closed container,
\(\left[ { PCl }_{ 5 } \right] =0.8\times { 10 }^{ -3 } \ mol \ { L }^{ -1 }\)
\( \left[ { PCl }_{ 5 } \right] =1.2\times { 10 }^{ -3 } \ mol \ { L }^{ -1 }\)
\(\left[ { Cl }_{ 2 } \right] \ =1.2\times { 10 }^{ -3 } \ mol \ { L }^{ -1 }\)
\( { K }_{ c } \ =\frac { \left[ { PCl }_{ 3 } \right] \left[ { Cl }_{ 2 } \right] }{ \left[ { { PCl }_{ 5 } } \right] }\)
\( =\frac { (1.2\times { 10 }^{ -3 })\times (1.2\times { 10 }^{ -3 }) }{ (0.8\times { 10 }^{ -3 }) } =1.8\times { 10 }^{ -3 }\)

The applications of dipole moment are
(a) The dipole moment helps to predict whether a molecule is polar or non-polar. As μ = q × d greater is the magnitude of dipole moment, higher will be the polarity of the bond. For non-polar molecules, the dipole moment is zero.
(b) The percentage of ionic character can be calculated as Percentage of ionic character \(=\frac { { \mu }_{ observed } }{ { \mu }_{ ionic } } \times 100\)
(c) Symmetrical molecules have zero dipole moment although they have two or more polar bonds.
(d) It helps to distinguish between cis and trans-isomers. Usually cis-isomer has higher dipole moment than trans-isomer.
(e) It helps to distinguish between ortho, meta and para-isomers. Dipole moment of para-isomer is zero. Dipole moment of ortho-isomer is greater than that of.

In p -block, when we move from left to right in a period, the acidic character of the oxides increases due to increase in electronegativity. e.g.
( i ) 2nd period
B₂O₃ < CO₂ < N₂O₃ acidic character increases.
( ii ) 3rd period
Al₂O₃ < SiO₂ < P₄O₁₀ < SO₃ < Cl₂O₇ acidic character increases.
on  moving down the group, acidic character decreases and basic character increaseas.e.g.
(a) Nature of oxides of 13 group elements 

Nature of oxides of 15 group elements
N₂O₅   P₄O₁₀    As₄O₁₀    Sb₄O₁₀   Bi₂O3
Strongly acidic   Moderately acidic  Amphoteric  Amphoteric Basic
Among the oxides of same element, higher the oxidation state of the element, stronger is the acid. e.g. SO₃ is a stronger is the acid than SO₂.
B₂O₃ is weakly acidic and on dissolution in water, ti forms orthoboric acid. Orthoboric acid does not act as a protonic acid ( it does not ionise ) but acts as a weak Lewis acid.
B₂O₃  + 3H₂O  \(\rightleftharpoons\)  2H₃BO₃
Boron trioxide                       Orthoboric acid
B ( OH )₃  +  H-----OH \(\longrightarrow\) [ B ( OH )₄ ]^- + H⁺
Al₂O₃ is amphoteric in nature. It is insoluble in water bur dissolves in alkalies and react with acids.
Al₂O₃  + 2NaOH \(\overset { \triangle }{ \longrightarrow } \) 2NaAlO₂       +              H₂O
Aluminiun trioxide                     Sodium meta ailuminate
Al₂O₃ + 6HCl \(\overset { \triangle }{ \longrightarrow }\) 2AlCl₃  + 3H₂O
Aluminium chloride
Tl₂O is as basic as NaOH due to its lower oxidation state ( +1 )
Tl₂O + 2HCl \(\longrightarrow \) 2TlCl + H₂O
P₄O₁₀ on reaction with water gives orthophosphoric acid.
P₄O10  +  6H₂O  \(\longrightarrow\) 4H₃PO₄
Phosphorus pentaxide                   Orthophosphoric acid
Cl₂O₇ is strongly acidic in nature and on dissolution in water, it gives perchloric acid.
Cl₂O₇  +  H₂O \(\longrightarrow\) 2HClO₄
Dichlorine heptoxide       Perchloric acid

Given that, mass of Zn = 32.65 g
1 mole of gas occupies =22.7 L volume at STP
Atomic mass of Zn = 65.3\(\mu\) 
The given equation is 
\(\underset { 65.3 \ g }{ Zn } +2HCL\rightarrow { ZnCL }_{ 2 }+\underset { 1mol=22.7 \ L \ at \ STP }{ H_{ 2 } } \)
From the above equation, it is clear that 65.3 gZn, when reacts with HCL, produces = 22.7 of \(H_{ 2 }\) at STP
\(\therefore\) 32.65 g Zn, when reacts with HCL, will produce
\(=\frac { 22.7\times 32.65 }{ 65.3 } =11.35 \ L \ of \ { H }_{ 2 } \ at \ STP\)