Chemistry

Let initial moles of reactant taken = n. Total moles obtained for benzene sulphonic acid = 0.6 n (with % yield = 60%)

Moles of benzene sulphonic acid before reaction II = 0.6n

Moles obtained for phenol (with % yield 50%) = 0.6 * 0.5 n = 0.3 n

So, overall % age yield of complete

reaction = $\frac{0.3n}{n}*100=30\mathrm{\%}$  

Oxidation of HCHO (a) will produce HCOOH, which is produced by ants.

$\mathrm{\%}Opticalpurity=\frac{Observedrotationofmixture*100}{rotationofpureenautiomer}$

$=\frac{+12.6}{+30}*100=42\mathrm{\%}$

  $2KMn{O}_4+5H_2{\stackrel{+3}{C}}_2{O}_4+3H_{2}S{O}_4\left(dil\right)\to K_{2}S{O}_4+2MnS{O}_4+10C{\stackrel{+4}{O}}_2+8H_{2}O$

Oxidation state of carbon changes from +3 to +4

$\therefore$ change in O. S is 1

Complex is   $\left[Co{\left(N{H}_3\right)}_{4}C{l}_2\right]Cl$

Primary valency = No. of ionisable species = 1

Solubility of CaF₂ = S mol/L

$S=\frac{2.34*10^{-3}}{0.1*78}=\frac{2.34}{78}*10^{-2}=3*10^{-4}mol/L$          

$K_{sp}\left(Ca{F}_2\right)=4S^3=4{\left(3*10^{-4}\right)}^3=108*10^{-12}$        

= 0.0108 * 10^-8 (mol/L)³

  $C{N}^{-},N{O}^{+}and{O}_2^{2+}$ have bond order = 3

$C_{P,m}=C_{v,m}+R$

$C_{v,m}=20.785-8.314=12.471J{K}^{-1}mo{l}^{-1}$

$n=\frac{5000}{12.471*200}=\frac{25}{12.471}\approx 2$

$K{O}_2,N{O}_2,Cl{O}_2,NO$ are paramagnetic.

On decreasing pressure of NO by a factor of '2' the rate of reaction decreases by a factor of '4'

$\therefore$ Order of reaction w.r.t 'NO' = 2