Ncert Solutions Chemistry Class 11th

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\

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

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

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

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

$\therefore$ 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.

In electro-refining, impure metal (blister copper) is used as an anode while precious metal like Au, Pt gets deposited as anode mud.

$\Delta H=3{\left(\Delta H_{comb}\right)}_{C_2{H}_2\left(g\right)}-{\left(\Delta H_{comb}\right)}_{C_6{H}_6\left(l\right)}$

$=-1300*3-\left(-3268\right)kJ/mole$

$=\left(-3900+3268\right)kJ/mole$

$=-632kJ/mole$

In photo-electric effect;

  $\frac{hc}{\lambda }=h{v}_0+\frac{1}{2}m{v}^2$             

  $\frac{6.63*10^{-34}*3*10^8}{500*10^{-9}}=\left(6.63*10^{-34}*4.3*10^{14}\right)+\frac{1}{2}*9.0*10^{-31}*v^2$            

  $v=\sqrt[]{\frac{\left(3.978*10^{-19}-2.85*10^{-19}\right)}{4.5*10^{-31}}}=\sqrt[]{25*10^{10}}=5*10^{5}m/s$             

Ans. = 5

Hydroxyapatite is the calcium minerals whose molecular formula is $3C{a}_3{\left(P{O}_4\right)}_2.Ca{\left(OH\right)}_2$ ${}_{}{{}_{}}_{}{}_{}$ or $C{a}_{10}{\left(P{O}_4\right)}_6{\left(OH\right)}_2.$ ${}_{}{{}_{}}_{}{}_{}$ During contact with water dissolved F ion, hydroxyapatite converted into ${}_{}{{}_{}}_{}{}_{}$ $3C{a}_3{\left(P{O}_4\right)}_2.Ca{F}_2$ .

$\Delta H_{vap}=41KJmo{l}^{-1}$

$\Delta H=\Delta E+\Delta nRT$      

$H_2{O}_{\left(\mathcal{l}\right)}\to H_2{O}_{vap}$        

$\Delta n=1$             

R = 8.3 JK^-1 mol^-1

T = 373 K

$\Delta$ E (Change of internal energy) = $\Delta H-\Delta nRT$  

$?\Delta E=41-\frac{1*8.3*373}{1000}kJmo{l}^{-1}=41-3.09kJmo{l}^{-1}=37.91kJmo{l}^{-1}$  

Nearest integer = 38.