Chemistry

$P_B^O=$ vapour pressure of benzene at 20°C = 70 torr

$P_M^O$ = vapour pressure of toluene at 20°C = 20 torr

Mixture is equimolar, X_B = 0.5 and X_M = 0.5

Total vapour pressure (P_T) = 70 * 0.5 + 20 * 0.5 = 45 torr

Mole fraction of benzene in vapour phase $\left(x_B^{\text{'}}\right)=\frac{P_B^O{X}_B}{P_T}=\left(\frac{70*0.5}{45}\right)torr$

= 0.777 = 77.7 * 10^-2 torr

Ans. = 78 (the nearest integer)

Hence, electronic configuration of CO²⁺ is $\left[Ar\right]3d^{7}4s^0$ . In complex ${\left[Co{\left(CN\right)}_6\right]}^{4-},$ given ligand CN^- is $\therefore$ strong hence, after pairing in d-subshell, total number of unpaired electron =

spin magnetic moment = $\sqrt[]{1\left(1+2\right)}=\sqrt[]{3}=1.73BM$ $=1.73BM\approx 2.0BM$  

For Ni²⁺, crystal field CFSE magnitude shows the magnitude of absorbed light. Following are the energy absorbed order

  ${\left[NiC{l}_4\right]}^{2-}<{\left[Ni\left(H_{2}O\right)\right]}^{2+}<{\left[Ni{\left(CN\right)}_4\right]}^{2-}$

Hence order of colour of compounds are.

${\left[NiC{l}_4\right]}^{2-}>{\left[Ni{\left(H_{2}O\right)}_6\right]}^{2+}>{\left[Ni{\left(CN\right)}_4\right]}^{2-}$

Both statement are correct for the glass body heating, but reason is not correct explanation during heating process of glass, constituents unit rupture of glass body and gives the edge smoothness.

All structure can produce -CHO functional group which gives +ve test of Tollen's reagent except structure (II), because if forms ketonic group after tautomerisation

$X\underset{H_{2}S{O}_4}{\overset{Conc}{\to }}$ Brown fumes + Brown ring test with FeSO₄/H₂SO₄

$x\underset{HCl}{\overset{H_{2}S}{\to }}Y\left(ppt\right)\underset{HN{O}_3}{\overset{Conc}{\to }}dissolved\underset{N{H}_{4}OH}{\overset{}{\to }}$ Deep blue colour solution

In cation analysis of Cu⁺ ions, precipitate formed is CuS on treating with H₂S and HCl which dissolved in HNO₃ and produced blue colour complex solution ${\left[Cu{\left(N{H}_3\right)}_4\right]}^{++}$ in NH₄OH. Brown fumes and brown ring performance given by nitrate sample. Hence salt is Cu (NO₃)₂.

In $Br{O}_4^{-},Br$ has +7 O.S, therefore it only acts as a oxidizing reagent, does not undergo disproportionation reaction

$6O{H}^{-}+C{l}^{-}\to Cl{O}_3^{-}+3H_{2}O+6e^{-}$

Mole of  $Cl{O}_3^{-}orKCl{O}_3=\frac{10}{122.6}=0.8456mol$

1 mole of $Cl{O}_3^{-}$  or KClO₃ produced by 6 faraday (F).

$\therefore 0.08456$ mole of $Cl{O}_3^{-}$ or KClO₃ produced by (6 * 0.08156) F

$\frac{l*t}{F}=6*0.8156$

$l=\frac{6*0.08156*965000}{10*60*60}A=1.311A\approx 1A$