Ncert Solutions Chemistry Class 11th

Ethylene glycol (HO – CH₂ – CH₂ – OH) and terephthalic acid

2700 kJ energy released from 180 gm (1 mole) of glucose

$\therefore$ 1 kJ energy released from $\frac{180}{2700}$ gm of glucose

$\therefore$ 10000 kJ energy released from   $\left(\frac{180\times 10000}{2700}\right)$ gm of glucose

Amount of glucose = $\frac{18000}{27}=666.66\approx 667gm$

Hydrogen peroxide reduces iodine to iodide ion is basic medium as;

$H_2{O}_2+2O{H}^{-}+l_2\to O_2+2l^{-}+2H_{2}O$

$N{a}_{2}O+H_{2}O\to 2NaOH$

w = 20 g

$V_{H_{2}O}=500mL$        

Mole of Na₂O=   $\frac{20}{62}$

$?$ 1 mole of Na₂O gives 2 mole of NaOH

$\therefore \left(\frac{20}{62}\right)$ mole of Na₂O gives $\left(2\times \frac{20}{62}\right)$ moles of NaOH

$=\frac{20}{31}mole$    

Molarity of NaOH solution

$=\frac{\frac{20}{31}\times 100}{500}M=\frac{20}{31}\times 2M=\frac{40}{31}M$

= 1.29 M

=  $12.9\times 10^{-1}M\approx 13\times 10^{-1}M$

Ans. = 13

PV = nRT             PV = constant (at constant T)

Pressure increases & volume decreases, PV remains constant at constant T.

B. O. D. value < 5 ppm for clean water and B.O.D value of polluted water 17 ppm.

Portland cement contains

Dicalcium silicate = 26%

Tricalcium silicate = 51%

Tricalcium aluminate = 11%

Hence major percentage is of tricalcium silicate

$\Delta H={\left(E_{act}\right)}_f-{\left(E_{act}\right)}_b$

= x – (x + y) = -y

              = -45 KJ/mol

              Ans. = 45

Applying : ${\left(n_1+n_2\right)}_{initial}={\left(n_1+n_2\right)}_{final}$  

Assuming the system attains a final temperature of T (Such that 300 < T < 60)

(Heat lost by N₂ of container l) = (Heat gained by N₂ of container II)

$n_{1}Cm\left(300-T\right)=n_{2}Cm\left(T-60\right)$  

$\left(\frac{2.8}{28}\right)\left(300-T\right)=\frac{0.2}{28}\left(T-60\right)$  

14 (300 – T) = T – 60

$P=\left(\frac{3}{28}\right)\times 8.31\times \frac{284}{3\times 10^{-3}}\times 10^{-5}bar=0.84287bar$  

$P=84.28\times 10^{-2}bar$

$\approx 84\times 10^{-2}bar$  

Minimum temperature at which reaction becomes spontaneous is,

  $T_{min}=\frac{\Delta H^0}{\Delta S^0}$             

$\Delta {H^o}_{min}=\left[\Delta H_f^o\left(Fe\right)+\Delta H_f^o\left(CO\right)\right]-\left[\Delta H_f^o\left(FeO\right)+\Delta H_f^o\left(C\right)\right]$

$=\left[0-110.5\right]-\left[-266.3+0\right]$

$=155.8KJ/mol$

$\Delta S°=\left[\Delta S°\left(Fe\right)+\Delta S^o\left(CO\right)\right]-\left[\Delta S°\left(FeO\right)+\Delta S°\left(C\right)\right]$

$=\left(27.28+197.6\right)-\left(57.49+5.74\right)J/molK$

$=161.65J/molK$

$T_{min}=\frac{155.8\times 10^3}{161.65}K=963.8K\approx 964K$