Ncert Solutions Chemistry Class 11th

Ionic radii of cations is smaller than anions, also more the positive charge less be the ionic radii and more the negative charge more be the ionic radii. Hence correct order of ionic radii is, $p^{3-}>S^{2-}>C{l}^{-}>K^{+}>C{a}^{2+}$

Millimoles of HCl = 200 × 0.2 = 40

Millimoles of NaOH = 300 × 0.1 = 30

Heat released =  $\frac{30}{1000}\times 57.1\times 1000J=1713J$

$\left[\begin{array}{l}\rho =\frac{m}{v}\\ m=\rho v\end{array}\right]$

Mass of solution = 500 × 1 g

= 500 g

Specific heat of water = 4.18 Jg^-1 K^-1

$\Delta T=\frac{q}{mc}$

$=\frac{1713}{500\times 4.18}°C$  

$=81.96\times 10^{-2}°C\approx 82\times 10^{-2}°C$

Ans. = 82

(1) Standard enthalpy of formation for alkali metal bromides becomes more negative on descending down the group.

(2) Standard enthalpy of formation for LiF is most negative among alkali metal fluorides.

(3) In case of Csl, lattice energy is less but Cs⁺ having less hydration energy due to which it

The Bond dissociation energy of D₂ is greater than H₂, therefore D₂ reacts slower than H₂.

$Unitof\frac{a{n}^2}{V^2}=atm$

$Unitofa=\frac{atm\times unitof{V}^2}{unitof{n}^2}$

$=\frac{atm\times {\left(d{m}^3\right)}^2}{mo{l}^2}$

= atm dm⁶ mol^-2

Number of water molecule

Gypsum :- CaSO₄.2H₂O  =>          2

Plaster of paris :- CaSO₄.    =>          0.5

Dead burnt plaster : CaSO₄          =>          0

  $F-Be-F$   $\mu =0$ $$

BF3 $\mu =0$

H2O $\mu \ne 0$ $$

NH3 $\mu \ne 0$ $$

CCl4 $\mu =0$ $$

HCl $\mu \ne 0$ $$

$T_1=27°C=300K$ T2 = 45° C = 45 + 273 = 318 K

P1 = 30 atmP2 =?

V is constant due to rigid tank\

$\therefore \frac{P_1}{T_1}=\frac{P_2}{T_2}$

$\frac{30}{300}=\frac{P_2}{318}\therefore P_2=\frac{1}{10}\times 318=31.8atm\approx 32atm$

$\Delta H_f^0=\left(+\Delta H_{sub}^0\right)+\left(+\Delta H_{IE}^0\right)+\left(+\frac{1}{2}\Delta H_{BDE}^0\right)+\left(\Delta H_{EA}^0\right)+\left(-\Delta H_{LE}^0\right)$

$\Delta H_{LE}^0=89.2+419+121.5-348.6+436.7kJmo{l}^{-1}$

$\therefore$ Lattice energy = 717.8

$\underset{\left(acid\right)}{H_{2}S}+\underset{Base}{H_{2}O}?H_3{O}^{+}+H{S}^{-}$

$\underset{\left(acid\right)}{H_{2}O}+\underset{\left(Base\right)}{N{H}_3}?N{H}_{4}OH$

With H₂S water acts like base and with NH₃ it acts like acid.