Class 12th

When red phosphorus is heated in a sealed tube at 803 K, a - black phosphorus is formed.

$A{g}^{+}\left(aq\right)+2N{H}_3\left(aq\right)?Ag{\left(N{H}_3\right)}_2^{+}\left(aq\right)$

t = 00.8 M $\left(\frac{a}{2}\right)M$                  

0 $=\infty$   5 * 10^-8 M   $\left(\frac{a}{2}-1.6\right)M$   0.8M

$k_f=\frac{\left[Ag{\left(N{H}_3\right)}_2^{+}\right]}{\left[A{g}^{+}\right]{\left[N{H}_3\right]}^2}$

$10^8=\frac{0.8}{\left(5*10^{-8}\right){\left(\frac{a}{2}-1.6\right)}^2}$

$\frac{a}{2}=2$

 Concentration of NH₃ added =   $\frac{a}{2}=2M$

 Volume of solution = 2L

 Moles of NH₃ added = 2 * 2 mol = 4 mol.

E = (50 N/s) sin (t – x/c)

Energy 5.5 * 10^-12 J., vol v =?

$u=\frac{1}{2}{\epsilon }_0{E}^2\to$ energy density

$?u=\frac{U}{v}\Rightarrow U=uv$

= $\frac{1}{2}{\epsilon }_0{E}^2.v$

$v=\frac{5.5*10^{-12}*2}{8.55*2500*10^{-12}}*10^{6}c{m}^3$

= 497

$\cong 500c{m}^3$

(a) $Calamine:ZnC{O}_3$  

(b) Malachite : $CuC{O}_3.Cu{\left(OH\right)}_2$  

(c) Siderite : FeCO₃

(d) Sphalerite : ZnS

Let molarity of KMnO₄ = x

$KMn{O}_4+FeS{O}_4\to F{e}_2{\left(S{O}_4\right)}_3+M{n}^{2+}$

n = 5         n = 1         Ferric sulphate

equivalent of KMnO₄ = equivalent of FeSO₄

5 * x * 10 = 1 * 0.1 * 10

x = 0.02 M

Strength = (0.02 * 158) = 3.16g/L

= 316 * 10^-2 g/L

Ans. = 316

$\underset{\left(z=93\right)}{Np}\Rightarrow 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^2$

$4d^{10}5p^{6}6s^{2}4f^{14}5d^{10}6p^{6}7s^{2}5f^{4}6d^1$

Total number of f-electrons = 14 + 4 = 18 electrons

Ans. = 18

Dumas method,

Moles of N in N, N-dimethylaminopentane (C₇H₁₇N)

$=\frac{57.5}{115}=0.5mole$

$C_7{H}_{17}N+\frac{45}{2}CuO\to 7C{O}_2+\frac{17}{2}{H}_{2}O+\frac{1}{2}{N}_2+\frac{45}{2}Cu$

$\frac{n_{CuO}}{\left(\frac{45}{2}\right)}=\frac{n_{c_7{H}_{17}N}}{1}$

$n_{CuO}=\left(\frac{45}{2}\right)*0.5mol$

= 11.25 mol

= 1125 * 10^-2 mol

Ans. = 1125

(a) Shows intra molecular H-bonding.

(b) Shows inter molecular H-bonding.

(c) It does not shows intermolecular H-bonding due to high steric hindrance at o-position of benzene ring.

Let mass of water initially = x gram

Mass of sucrose = (1000 – x) gram

Mole of sucrose = $\frac{1000-x}{342}mol$

$?$ 0.75 molal Þ 0.75 mole solute in 1 kg of solvent

$0.75=\frac{\left(1000-x\right)/342}{x/1000}\left[molality=\frac{n_{solute}}{M_{solvent}\left(kg\right)}\right]$

$4=\frac{\frac{\left(1000-795.86\right)*1.86}{342}}{a}$        

a = 0.2775kg or 277.5 gram

Ice separated = (795.86 – 277.5) gram

= 518.3 gram

Ans. = 518 (the nearest integer)

In the lyophilic colloids, the colloidal particles are extremely solvated.