Chemistry

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

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.

Millimoles of HCl = 200 * 0.2 = 40

Millimoles of NaOH = 300 * 0.1 = 30

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

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

Mass of solution = 500 * 1 g

= 500 g

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

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

$=\frac{1713}{500*4.18}°C$  

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

Ans. = 82

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.

mvr = $\frac{nh}{2\pi }\left(angularmomentum\right)$  

$KE=\frac{n^2{h}^2}{8{\pi }^{2}m{r}^2}$ (Bohr's kinetic energy)

$=\frac{4h^2}{8{\pi }^{2}m{\left(4a_0\right)}^2}\left(n=2\right)$

$K.E.=\frac{1}{32{\pi }^2}\frac{h^2}{m{a}_0^2}$

Comparing with $\frac{h^2}{xm{a}_0^2}$

x = 32π² = 315.50

10x = 3155

(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 is less soluble in water.

(4) For alkali metal fluorides, the solubility in water increases from Li to Cs. LiF is least soluble in water.

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.