Ncert Solutions Chemistry Class 11th

Due to higher extent of polarization by Li⁺ and Mg²⁺, LiCl and Mgcl₂ have covalent character. Therefore they are soluble in ethanol.

Due to very high value of lattice energy, LiF is having very less solubility in water.

The highest industrial consumption of hydrogen gas is in the synthesis of ammonia gas (Having manufacturing of N-based fertizers)

The highest industrial consumption of hydrogen gas is in the synthesis of ammonia gas (Having  manufacturing of N-based fertizers)

% 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.