Ncert Solutions Chemistry Class 11th

% of C in organic compound

$=\frac{12}{14}*\frac{WC{O}_2}{W.O.C.}*100$  

$=\frac{952.56}{21.648}=44\mathrm{\%}$  

$\mathrm{\%}ofH=\frac{2}{18}*\frac{W{H}_{2}O}{W.O.C.}*100$  

= $\frac{20}{18}*\frac{0.4428}{0.492}*100$  

=  $\frac{88.56}{8.856}=10\mathrm{\%}$  

= 100 – 54 = 46%

In 4d orbital, n = 4 and $\mathcal{l}=2$

Radial nodes = $n-\mathcal{l}-1$

Radial nodes = 4 – 2 – 1 = 1

And angular nodes,  $\mathcal{l}=2$

x      +      y    +   3z     =     xyz₃

1 mole 1 mole 0.05 mole

$?n_x=1$

$n_y=1$

$n_z=\frac{0.05}{3}=0.0167$ here z is limiting reagent.

$?\frac{0.05}{3}$ mole z gives 1 mole xyz₃

$?$ mass of xyz₃ = n * molecular mass

 = $\frac{0.05}{3}*\left(10+20+3*30\right)a.m.u.$  

= 0.5 * 4 = 2g

Acidic oxide -> Cl₂O₇

Neutral oxide -> N₂O, NO

Basic oxide ->Na₂O

Amphoteric oxide -> As₂O₃

$2C\left(s\right)+2H_2\left(g\right)\text{?}\text{?}\to C_2{H}_6\left(g\right)\Delta H=\Delta H_f\left(C_2{H}_6\right)$  

$\Delta H_f\left(C_2{H}_6\right)=\left[2*\left(-394\right)\right]+\left[3*\left(-286\right)\right]-\left[1*\left(-1560\right)\right]KJ/mole$  

= (-788 – 858 + 1560) KJ/mole = -86 kJ/mole

Number of moles of C = Number of moles of CO₂ = $\frac{330}{44}$ moles

Number of moles of H = 2 * no. of moles of H₂O = $\left(\frac{270}{18}*2\right)$ moles

Mass of C =  $\frac{330}{44}*12$ gm = 90 gm

Mass of H =  $\frac{270}{18}*2*1gm=30gm$

$\begin{array}{l}\mathrm{\%}ofC=\frac{90}{120}*100\mathrm{\%}=75\mathrm{\%}\\ \mathrm{\%}ofH=\frac{30}{120}*100\mathrm{\%}=25\mathrm{\%}\end{array}$  

Strength = 0.72 g/L

Using ;

$Molarity=\frac{Strength}{Molarmassofsolute}$              

$M=\frac{0.72}{180}M$              

= 4 * 10^-3 M

Ans. is 4.

Reduced pressure distillation technique is used for purification of high boiling organic liquid which decompose near its boiling point. By reducing pressure they boil below their normal boiling point.

Solubility of gas in liquid increases on decreasing the temperature. So water at 4°C will have more dissolved O₂.

Ba has outer electronic configuration 6s².

CaC₂O₄ is insoluble in water.

Compound of Li are covalent so soluble in organic solvent.

Na forms strong monoacidic base.