66 - General Chemistry Questions Answers

 Joshi sir comment

Constant b is associated to volume correction so its value be less for molecule of smaller volume.

Sharing is the case applicable for the atoms of same electronegativity, thus by sharing electrons (charges) atoms make their orbital configurations stable according to the rules defined for stability.

In first case reactions involved are$$\begin{gathered} \mathrm{Reduction} \mathrm{of} {\mathrm{Fe}}^{+++}, {\mathrm{Fe}}^{+++}\to {\mathrm{Fe}}^{++} \\ \mathrm{Titration} \mathrm{step}, 6{\mathrm{Fe}}^{++}+{\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7\to 6{\mathrm{Fe}}^{+++}+2{\mathrm{Cr}}^{+++} \\ \mathrm{By} \mathrm{using} \mathrm{titration} \mathrm{reaction}, \mathrm{solve} \mathrm{the} \mathrm{question} \end{gathered}$$

 In second case reaction involved is${\mathrm{As}}_4{\mathrm{O}}_6+4{{\mathrm{I}}_3}^{-}\to {\mathrm{As}}_4{\mathrm{O}}_{10}+12{\mathrm{I}}^{-}$Now solve

inform if any issue

by formula PV/T = constant

(752-28)*150/301 = 760*V/273

solve for V, this will be the equivalent volume at STP

now convert this volume in gm by the formula 2V/22400

then finally m = zit, where 

z = 1/96500

solve it for i

by gas eq.  PV/nT = constant

so 0.5*2V/0.7*300 = PV/n₁400 + PV/n₂300        (1)

no. of moles are also constant

so 0.7 = n₁ + n₂

similarly by gas eq. 

PV = n₁R400                      for first flask

PV = n₂R300                      for second flask

now solve

b = 4V, where V is volume of 1 mole particles

so 24*10^-6 cm³/mole = 4*[4πr³/3] cm³/mole 

now solve

= 3RT/2 

3FeO.1Fe₂O₃ = Fe₅O₆ = Fe_5/6O