Chemistry

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.

$O_2\left(g\right)\underset{}{\overset{UV}{\to }}O\left(g\right)+O\left(g\right)$  

$O\left(g\right)+O_2\left(g\right)\stackrel{UV}{\to }{O}_3\left(g\right)$      

Ozone in the stratosphere is a product of UV radiations acting on O₂ molecule.

$2K_{2}C{r}_2{O}_7+8H_{2}S{O}_4+3C_2{H}_{6}O\to 2C{r}_2{\left(S{O}_4\right)}_3+3C_2{H}_4{O}_2+2K_{2}S{O}_4+11H_{2}O$

Given that, rate of appearance of Cr₂(SO₄)₃ = $+\frac{d\left[C{r}_2{\left(S{O}_4\right)}_3\right]}{dt}=2.67mole/min$  

LHS                                                                 RHS

$\left(\frac{Rateofdisappearanceof{C}_2{H}_{6}O}{3}\right)=\left(\frac{RateofappearanceofC{r}_2{\left(S{O}_4\right)}_3}{2}\right)$            

Rate of disappearance of C₂H₆O = $\frac{2.67*3}{2}mol/min$  

Rate of disappearance of C₂H₆O = 4.005 mol /min

Ans. = 4 (nearest integer)

Mass of organic compound = 0.2 gram

$n_{AgBr}=\frac{0.188}{188}mol$

= 0.001 mol

$n_{Br}=n_{AgBr}=0.001mole$

 Mass of Br = 0.001 * 80g

= 0.08 g

Mass & of Br = $\frac{0.08}{0.2}*100\mathrm{\%}$

= 40 %

$2KMn{O}_4\stackrel{\Delta }{\to }\underset{\left(green\right)}{\overset{+6}{K_{2}Mn{O}_4}}+\underset{\left(black\right)}{Mn{O}_2}+O_2$

Mn⁶⁺ has one unpaired electron so paramagnetic and has green colour.