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

The phenomenon of existence of an element in two or more forms which differ in physical properties but have almost same chemical nature is known as allotropy and the different forms of the element are known as allotropes.

Molar mass of \({ H }_{ 2 }{ O }_{ 2 }=34 \ gmol^{ -1 }\)
1L of 5M solution of \({ H }_{ 2 }O_{ 2 }\)will contain 34 x 5g \({ H }_{ 2 }O_{ 2 }\)
2L of 5M solution of \({ H }_{ 2 }O_{ 2 }\)will contain 34 x 5 x 2 = 340g \({ H }_{ 2 }O_{ 2 }\)
Mass of \({ H }_{ 2 }O_{ 2 }\)present in 2L of 5 molar solution = 340g

\(BaS{ O }_{ 4 }(S) \ \leftrightharpoons \ { Ba }^{ 2+ }(aq)+{ SO }_{ 4 }^{ 2- }(aq)\)
\({ K }_{ sp } \ for \ BaS{ O }_{ 4 }=\left[ { Ba }^{ 2+ } \right] \left[ { SO }_{ 4 }^{ 2- } \right] =S\times S={ S }^{ 2 }\)
\( But \ S=8\times { 10 }^{ -4 }mol \ { dm }^{ -3 }\)
\(\therefore { K }_{ sp }={ \left( 8\times { 10 }^{ -4 } \right) }^{ 2 }=64\times { 10 }^{ -8 }\)
   In the presence of 0.001 MH₂SO₄ , the expression for K_sp will be
\({ K }_{ sp }=\left[ { Ba }^{ 2+ } \right] \left[ { SO }_{ 4 }^{ 2- } \right]\)
\( { K }_{ sp }=(S).(S+0.01)\)
\((0.01 \ M \ { SO }_{ 4 }^{ 2- } \ Iions \ from \ 0.01 \ M \ { H }_{ 2 }{ SO }_{ 4 })\)
\(\Rightarrow \ 64\times { 10 }^{ -8 }=S.(S+0.01)\)
\(\Rightarrow { S }^{ 2 }+0.01S-64\times { 10 }^{ -8 }=0\)
\(\therefore \ S=\frac { -0.01\pm \sqrt { { \left( 0.01 \right) }^{ 2 }+{ \left( 4\times 64\times { 10 }^{ -8 } \right) } } }{ 2 }\)
\(=\frac { -0.01\pm \sqrt { { 10 }^{ -4 }+{ \left( 256\times { 10 }^{ -8 } \right) } } }{ 2 }\)
\( =\frac { -0.01\pm \sqrt { { 10 }^{ -4 }+{ \left( 1+256\times { 10 }^{ -4 } \right) } } }{ 2 } \)
\(=\frac { -0.01\pm { 10 }^{ -2 }\sqrt { 1+0.0256 } }{ 2 } =\frac { { 10 }^{ -2 }(-1\pm 1.012719) }{ 2 } \)
\( =5\times { 10 }^{ -3 }(-1+1.012719)=6.4\times \ mol \ { dm }^{ -3 }\)

In general, if enthalpy of an overall reaction A\(\rightarrow \) B along one route is  \(\Delta _{ r }H \ and \ \Delta _{ r }H_{ 1 },\Delta _{ r }H_{ 2 },\Delta _{ r }H_{ 3 }....\) representing enthalpies of reactions leading to same product B along another route, then we have
\(\Delta _{ r }H=\Delta _{ r }H_{ 1 }+\Delta _{ r }H_{ 2 }+\Delta _{ r }H_{ 3 }+....\)
Note
For a general reaction Hess's law of constant heat summation can be represented as

\(N_{ 2 }{ O }_{ 4 }(g)\leftrightharpoons 2NO_{ 2 }(g)\)
If N₂O₄ is 50% dissociated,d, the mole fraction of both the substances is given by
\(x_{ N_{ 2 }O_{ 4 } }=\frac { 1-0.5 }{ 1+0.5 } \Rightarrow x_{ NO_{ 2 } }=\frac { 2\times 0.5 }{ 1+0.5 }\)
\( p_{ N_{ 2 }O_{ 4 } }=\frac { 0.5 }{ 1.5 } \times 1atm,\ p_{ NO_{ 2 } }=\frac { 1 }{ 1.5 } \times 1atm\)
The equilibrium constant K_p is given by
\(K_{ p }=\frac { (p_{ NO_{ 2 } })^{ 2 } }{ p_{ { N }_{ 2 }{ O }_{ 4 } } } =\frac { 1.5 }{ (1.5)^{ 2 }(0.5) } =1.33atm\)
Since,
\(\Delta _{ r }G^{ \circ }=-RT \ ln \ K_{ p }\)
\( \Delta _{ r }G^{ \circ }=(-8.314 \ JK^{ - } \ mol^{ - })\times (333K)\times (2.303)\times (0.1239)\)
\( =-763.8 \ kJmol^{ -1 }\)

Ionic bond The bond formed, as a result of the electrostatic between the positive and negative ions was termed as the electrovalent bond or ionic bond. e.g. the formation of NaCL from sodium and chlorine can be explained as
\(Na\longrightarrow { Na }^{ + }+{ e }^{ - }\)   
\(\left[ Ne \right] { 3s }^{ 1 } \ \left[ Ne \right] \)
\(Cl \ + { e }^{ - }\longrightarrow { Cl }^{ - }\)
\(\\ \left[ Ne \right] { 3s }^{ 2 }{ 3p }^{ 5 } \left[ Ne \right] { 3s }^{ 2 }{ 3p }^{ 6 }or\left[ Ar \right] \)
\({ Na }^{ + }+{ Cl }^{ - }\longrightarrow NaCl \ or \ { Na }^{ + }{ Cl }^{ - }\) 
Similarly, the formation of CaF₂ may be shown as
\( Ca \longrightarrow { Ca }^{ 2+ } +{ 2e }^{ - }\)    
\(\left[ Ar \right] { 4s }^{ 2 } \left[ Ar \right] \)
\(F \ +\ { e }^{ - }\longrightarrow \ { F }^{ - }\) 
\(\\ \left[ He \right] { 2s }^{ 2 }2p^{ 5 } \left[ He \right] { 2s }^{ 2 }2p^{ 6 }or \ \left[ Ne \right] \)
\({ Ca }^{ 2+ }+{ 2F }^{ - }\longrightarrow Ca{ F }_{ 2 } \ or \ { Ca }^{ 2+ }{ 2F }^{ - } \)  
Covalent bond The bond formed between the two atoms by mutual sharing of electrons between them is called covalent bond. e.g. the formation of chlorine molecules can be explained

Similarly, in the formation of HCl
    

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.

All the elements of group IA (or I) i.e. alkali metals have the similar outer electronic configuration i.e. ns¹ where n refers to the number of principle shells. Their electronic configurations are given below.

Hence, placement of all these elements of group 1 of the periodic table because of similarity in electronic configuration and all the elements have similar properties. Thus, it is also evident that the physical and chemical properties of the elements depend on their atomic number and not on their atomic mass.
Thus, it is also evident that the physical and chemical properties of the elements depend on their atomic number and not in their atomic mass.