Class 11th

23 * 3 gm of Na contained by 164 gm of Na₃PO₄

So, 3.45 gm of Na contained by $\left(\frac{164*3.45}{69}\right)$ gm of Na₃PO₄

Therefore mole of Na₃PO₄= $\frac{164*3.45}{69*164}$ = 0.05 mol

Molarity = $\frac{0.05*1000}{100}$ M = 0.5 M

= 50.0 * 10^-2 M

For equation A + B $?C+D$  

  $K_C=\frac{\left[C\right]\left[D\right]}{\left[A\right]\left[B\right]}=100at298K$             

Here;    A +  B  $?$ C    +    D

t = 0       1M        1M

eq.         (1-x)      (1-x)      (1-x)      (1-x)

$Kc=\frac{\left(1+x\right)*\left(1+x\right)}{\left(1-x\right)*\left(1-x\right)}$

$\therefore x=\frac{9}{11}$  

At equilibrium, concentration of D is = 1 + $\frac{9}{11}=\frac{11+9}{11}=\frac{20}{11}$  

= 1.818 = 181.8 * 10^-2 M

Ans. = 182 (the nearest integer)

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

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

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

$=-632kJ/mole$

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 ${}_{}{{}_{}}_{}{}_{}$ $3C{a}_3{\left(P{O}_4\right)}_2.Ca{F}_2$ .

According to Bohr's model, velocity of electron increases according to atomic number (z).

$v_e\propto \frac{z}{n}$ (z = Atomic number and n = number of shell/principal quantum number)

Similarly, velocity of electron decreases according to number of shell (i.e. principal quantum number).

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

$\Delta H=\Delta E+\Delta nRT$      

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

$\Delta n=1$             

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.

Statement I is true because methyl orange indicator has pH range (3.2 to 4.2) which is suitable for buffer produced by strong acid and weak base.

Statement II is false because phenolphthalein has pH range (8.4 to 10.2) and only suitable for equivalence point above than 7, In this case pH increased by the addition of NaOH on acetic acid.

De-ionised water means dissolved minerals free water which is prepared by synthetic resin method. Where cation exchanged by H⁺ and anion exchanged by OH^- of resin.

$Ca{\left(OH\right)}_2+C{O}_2\to CaC{O}_3+H_{2}O$ (1st step of reaction)

$CaC{O}_3+H_{2}O+C{O}_2\to Ca{\left(HC{O}_3\right)}_2$ (2nd step of reaction)