The two cells may be represented as
Zn I Zn²⁺(cone = c) II Cu²⁺(c = ?) I Cu, EMF = E₁
Zn I Zn²⁺(cone = c) II Cu²⁺(0·5 M) I Cu, EMF = E₂
The cell reaction is : Zn + Cu²⁺ ⇾ Zn²⁺+ Cu
\(E_1=E^0-{2.303\ RT\over 2F}log{c\over [Cu^{2+}]}\)
\(E_2=E^0-{2.303\ RT\over 2F}log{c\over 0.5}\)
\(E_2-E_1={2.303\ RT\over 2F}\left( log{c\over [Cu^{2+}]}-log{c\over 0.5}\right)=0.03V\)
\({0.06\over 2}log{0.5\over [Cu^{2+}]}=0.03\ or\ log{0.5\over [Cu^{2+}]}=1\ or\ {0.5\over {[Cu^{2+}]}}=10\ or\ [Cu^{2+}]={0.5\over 10}=0.05M\)

Assume that we have 100 g of solution (one can start with any amount of solution because the results obtained will be the same). Solution will contain 20 g of ethylene glycol and 80 g of water.
Molar mass of C₂H₆O₂ = 12 x 2 + 1 × 6 + 16 × 2 = 62 g mol^-1
\(\text {Moles of } \mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}=\frac{20 \mathrm{~g}}{62 \mathrm{~g} \mathrm{~mol}^{-1}}=0.322 \mathrm{~mol}\)
\(\text {Moles of water }=\frac{80 \mathrm{~g}}{18 \mathrm{~g} \mathrm{~mol}^{-1}}=4.444 \mathrm{~mol}\)
\(\mathrm{x}_{\text {glycol }}=\frac{\text { moles of } \mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}}{\text { moles of } \mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}+\text { moles of } \mathrm{H}_{2} \mathrm{O}}\)
\(=\frac{0.322 \mathrm{~mol}}{0.322 \mathrm{~mol}+4.444 \mathrm{~mol}}=0.068\)
\(\text {Similarly, } x_{\text {water }}=\frac{4.444 \mathrm{~mol}}{0.322 \mathrm{~mol}+4.444 \mathrm{~mol}}=0.932\)
Mole fraction of water can also be calculated as: 1- 0.068 = 0.932

(i) For a solution of volatile liquids, at a given temperature the partial vapour pressure of each component in solution is equal to the product of vapour pressure of the pure component and its mole fraction.
(ii) Henry's law states that the mass of a gas dissolved per unit volume of the solvent at a constant temperature is directly proportional to the pressure of the gas in equilibrium with the solution.

Osmotic pressure is defined as the excess pressure which must be applied to a solution to prevent the passage of solvent into it through the semi-permeable membrane.
Osmotic pressure \((\pi )\) at temperature (T) is
 \(\pi =\frac { n }{ V } RT,\)
Determination of molecular mass from osmotic pressure
According to Van't Hoff equation,
\(\pi =\frac { n }{ V } RT,\)
where n is the number of moles of the solute and may be given as  \(\frac { { \omega }_{ B } }{ { M }_{ B } } .\) Here  \({ \omega }_{ B }\) is the weight of the solute and \({ M }_{ B }\) is its molecular mass. Substituting the value of n in the above expression, we get:
\(\pi =\frac { { \omega }_{ B }RT }{ { M }_{ B }V } or \ { M }_{ B }=\frac { { \omega }_{ B }RT }{ V\pi } \)
Thus, the molecular mass of the solute, \({ M }_{ B }\) can be calculated.

Due to higher pressure exerted by steam than in the open pan.

01 M BaCI₂ solution will have higher boiling point because of large number of particles on dissociation.
\( \mathrm{NaCI} \longrightarrow \mathrm{Na}^{+}+\mathrm{CI}^{-}(2 \text { particles }) \\ \mathrm{BaCI}_2 \longrightarrow \mathrm{Ba}^{2+}+2 \mathrm{CI}^{-}(3 \text { particles }) \)

Below 1973 K, the Δ_fG^o curve for the formation of SiO₂ lies above the.Δ_fG^o curve for the formation of MgO, therefore, at temperatures below 1973 K, Mg can reduce SiO₂ to metallic silicon
\(SiO_2+2Mg\xrightarrow{<1973K}2MgO+Si;Δ_rG^0=-ve\)
Above 1973 K, the ∆ _fG⁰ curve for the formation of SiO₂ lies below the corresponding curve for the formation of MgO. Therefore, above 1973 K, silicon can reduce MgO to Mg
\(Si + 2MgO \xrightarrow{>1973K}SiO_2+2Mg\)

The activation energy for the reaction C (diamond) \(\longrightarrow\) C (graphite) is very high which is not available at room temperature.

Suppose B is in excess so that b>> a or x. Neglecting a and x in comparison to b, the equation reduced to k b = k' = \(\frac {2.303}{t} log \frac {a}{(a-x)}\) which is same as for reactions of 1st order.

The molar conductivity of a solution at infinite dilution, i.e., when concentration approaches zero is called limiting molar conductivity. With decrease in concentration, i.e., with dilution, number of ions per cm³ decreases due to large increase in volume of the solution. As conductivity is the conductance per cm³ of the solution, hence it decreases.

In concentrated solution of a strong electrolyte, the interionic forces of attraction are large.

258 g mol^-1

\({ p }_{ X }^{ 0 }\)= 600mm

\({ p }_{ Y }^{ 0 }\)= 400mm

NaCI, being an electrolytes, dissociates almost completely to give Na⁺ and CI^- ions whereas glucose, being non - electrolyte, does not dissociate. Hence, the number of particles in 0.1 M NaCI solution is nearly double than in 0.1 M glucose solution. Freezing point depression, being a colligative property, is therefore, nearly twice for NaCI solution than for glucose solution of same molarity.

Water as ice

We know that, m = K_H \(\times\) P
\(\therefore\) 6.56 \(\times\) 10^-2g = K_H \(\times\) 1 bar ...(i)
\(\therefore\) 0 5.00 \(\times\) 10^-2 g = K_H \(\times\) P.........(ii)
K_H = 6.56 \(\times\) 10^-2/1 bar (from i)
K_H = 5.00 \(\times\) 10^-2//p bar (from ii)
\(\therefore \quad \frac{6.56 \times 10^{-2}}{1}=\frac{5.00 \times 10^{-2}}{p}\)
\(\therefore \quad P=\frac{5 \cdot 00}{6 \cdot 56}=0.762 \text { bar. }\)

After time, t, total pressure, \(P_{t}=\left(P_{0}-p\right)+p+p\) 
\(\Rightarrow P_{t}=P_{0}+p\) 
\(\Rightarrow p=P_{t}-P_{0}\) 
Therefore,\(P_{0}-p=P_{0}-\left(P_{t}-P_{0}\right)\) 
= 2P₀ − P_t
For a first order reaction,
\(=\frac{2.303}{t} \log \frac{P_{0}}{P_{0}-p}\) 
\(=\frac{2.303}{t} \log \frac{P_{0}}{2 P_{0}-P_{t}}\) 
When t = 360,\(k=\frac{2.303}{360 s} \log \frac{35.0}{2 \times 35.0-54.0}\) 
= 2.175 × 10⁻³ s⁻¹
When t = 720 s, \(k=\frac{2.303}{360 s} \log \frac{35.0}{2 \times 35.0-54.0}\) 
= 2.175 × 10⁻³ s⁻¹
When t = 720 s, \(\mathrm{K}=\frac{2.303}{720 \mathrm{~s}} \log \frac{35.0}{2 \times 35.0-63.0}\) 
= 2.235 × 10⁻³ s⁻¹
Hence, the average value of rate constant is
\(k=\frac{\left(2.175 \times 10^{-3}\right)+\left(2.235 \times 10^{-3}\right)}{2} s^{-1}\) 
= 2.21 × 10⁻³ s⁻¹

Calculation of activation energy of reaction (i)
T₁= 300 K, T₂ = 310 K, k₁ = k, k₂ = 2 k
\(log{k_2\over k_1}={E_A\over 2.303E}\left(T_2-T_2\over T_1T_2\right ),ie.,\ log2={E_a\over 2.303\times8.314}\times{10\over 300\times310}\ or\ E_a=53.60kJmol^{-1}\)
Calculation of rate constant of reaction (i) at 310 K
\(k={0.693\over t_{1/2}}={0.693\over 30\ min}=2.31\times10^{-2}min^{-1}\)
Rate constant of reaction (ii) at 310 K = 2 x 2·31x 10^-2 min^-1 = 4·62 x 10^-2 min^-1
Energy 0f ac tiva tion 0f reac tion (ii) =\({53.60kJ\ mol^{-1}\over 2}=26.80kJ\ mol^{-1}\)
Aim. To calculate k for reaction (ii) at 300 K
\(log{4.62\times10^{-2}\over k_{300}}={26.80\over 2.303\times8.314\times10^{-3}}\times{10\over 300\times310}=0.0151\)
or \({4.62\times10^{-2}\over k_{300k}}=Antilog\ or\ 0151 = 1.035\ or \ k_{300k}={4.62\times10^{-2}\over 1.035}=4.46\times10^{-2}min^{-1}\)

2.88 \(\times\) 10^-4 cm

The choice of reducing agent and temperature
is decided with the help of Ellingham
diagram and sign of Δ_rG_o. AI can reduce
MgO at 2000 K because at this temperature
range, the line for Δ_fG^o (Mg, MgO)
lies above the line Δ_fG^o (A_I, AI₂P³). This
indicates that at this temperature, AI₂P₃ is
more stable than MgO and Δ_rG^o for the
reaction given below would be negative
above 1665 K. Δ_rG^o is standard free energy
of reaction and Δ_fG^o is standard free
energy of formation of a compound.
2AI + 3MgO ➝ Δ_rG^o + 3Mg
Below 1665 K, Mg can reduce AI₂P₃ i to AI, because Δ_fG^o will be negative for the reaction given below.
Since Δ_rG^o is negative at a particular temperature, therefore, co is suitable reducing agent for Fe₂O₃.The temperature at which reduction will take place can be calculated with the help of Ellingham diagram.

(i) It is done so as to reduce its melting point and increases electrical conductivity.
(ii) It is because 'Cr' is stronger reducing agent than 'Zn'.
(iii) It is because it is easier to reduce oixde to get metal than from sulphide. \(\Delta G\) is more negative in the case of reduction of oxide than reduction of sulphide directly. 

Body temperature of the human body remains constant. Low concentration of oxygen in the blood at altitude is due to low atmospheric pressure.

Camphor has a very high cryoscopic constant (=39.7^o). Hence, It gives a large depression in melting point when an organic solute is dissolved in it.

\(\pi =CRT\\ C=0.002\quad mol\quad { dm }^{ -3 }=2\quad mol\quad { m }^{ -3 }\\ R={ 8.314JK }^{ -1 }{ mol }^{ -1 },T=293K\\ \pi =2\times 80314\times 293=4872\quad Pa\)

On increasing the temperature, volume of the solution increases. Hence Molarity decreases.