Chemistry

Mass of C₁₅H₃₀ = volume * Density

= 1000 $m\mathcal{l}$  * 0.756 gm/ $m\mathcal{l}$  

C₁₅H₃₀ + 22.5 O₂   $\stackrel{}{\to }$ 15CO₂ + 15H₂O

No. of moles of C₁₅H₃₀ = $\frac{756}{210}$ moles

No. of moles of O₂ required =   $\left(22.5*\frac{756}{210}\right)moles$

Mass of O₂ required = 22.5 *   $\frac{756}{210}*32=2592gm$

No. of moles of CO₂ liberated = 15 *   $\left(\frac{756}{210}\right)$ moles

Mass of O₂ liberated =   $15*\frac{756}{210}*44=2376gm$

n-factor of KMnO₄ in acidic medium = 5

n-factor of Mohr's salt = 1

meq of KMnO₄ = meq of Mohr's salt

0.01 * 5 * V = 0.05 * 1 * 20

Volume of KMnO₄ used, V = 20 mL

So; Volume of KMnO₄ left in burette = 50 -20mL

= 30 mL

Moles of C = moles of CO₂

=  $\frac{0.793}{44}mol$

Mass of C =  $\frac{0.793}{44}*12g$

= 0.261g

Moles of H = 2 *  $\frac{0.442}{18}mol$

Mass of H =  $2*\frac{0.442}{18}*1g$

Total mass of compound = 0.492g (given)

So; mass of O = (0.492 – 0.216 – 0.049) g

= 0.227g

% of O = $\frac{0.227}{0.492}*100$  

= 46.14%

the nearest integer = 46

Bromination through free radical mechanism occurs at allylic carbon.

(a) $CoC{\mathcal{l}}_3.4N{H}_{3}is\left[CoC{l}_2{\left(N{H}_3\right)}_4\right]Cl$ will show cis- trans isomerism as;

(b) $CoC{\mathcal{l}}_3.5N{H}_{3}is\left[Co{\left(N{H}_3\right)}_{5}C\mathcal{l}\right]C{\mathcal{l}}_2$ will not show cis- trans isomerism.

(c) $CoC{\mathcal{l}}_3.6N{H}_{3}is\left[Co{\left(N{H}_3\right)}_6\right]C{\mathcal{l}}_3$ will not show cis- trans isomerism.

(d) $CoC\mathcal{l}{\left(N{O}_3\right)}_2.5N{H}_{3}is\left[Co{\left(N{H}_3\right)}_{6}C\mathcal{l}\right]{\left(N{O}_3\right)}_2$  will not show cis-trans isomerism.

CN is a strong field ligand, so pairing occurs.

1. ${\left[Fe{\left(CN\right)}_6\right]}^{4-}$

So; it is diamagnetic

2. ${\left[Fe{\left(CN\right)}_6\right]}^{3-}$

So; it is paramagnetic.

3. ${\left[Ti{\left(CN\right)}_6\right]}^{3-}$

Ti³⁺ = 4s⁰3d¹

 It is paramagnetic

4. ${\left[Ni{\left(CN\right)}_4\right]}^{2-}$

It is diamagnetic

5. ${\left[Co{\left(Cn\right)}_6\right]}^{3-}$

Hence,  ${\left[Fe{\left(CN\right)}_6\right]}^{3-}and{\left[Ti{\left(CN\right)}_6\right]}^{3-}$ ${{}_{}}^{}{{}_{}}^{}$ are paramagnetic.

$t_{1/2}=200days=\frac{0.693}{k}$

$t=\frac{2.303}{k}lo{g}_{10}\frac{N_0}{N}$

83 =  $\frac{2.303}{0.693}*200log\frac{N_0}{N}$

$83=\frac{200}{0.3010}log\frac{N_0}{N}$

0.125 = $log\frac{N_0}{N}$

$\frac{N_0}{N}=antilog\left(0.125\right)$

= 1.333

Activating remaining = $\frac{N}{N_0}*100$

$=\frac{\frac{N_0}{1.333}}{N_0}*100$

= 75%

$\pi =\frac{n}{v}RT$

$\pi =\frac{W}{M*v}RT$

$5.03*10^{-3}=\frac{2.5}{M*0.5}*0.083*300$

$M=\frac{2.5*0.083*300}{5.03*0.5}*10^{3}g/mol$

Molar mass of protein, M = 24751 g/mol

 Molar mass of glycine = 75 g/mol

So; number of glycine units =   $\frac{24751}{75}=330$

$E_{S{n}^{2+}/Sn}^0=-0.140V=E_2^0$

$E_{S{n}^{4+}/Sn}^0=+0.010V=E_3^0$

Here, $\Delta G_3^0=\Delta G_1^0+\Delta G_2^0$

$-4F{E}_3^0=-2F{E}_1^0-2F{E}_2^0$

$E_3^0=\frac{E_1^0+E_2^0}{2}$

$+0.010=\frac{E_1^0+\left(-0.140\right)}{2}$

$E_1^0=E_{S{n}^{4+}/S{n}^{2+}}^0=+0.16$

= 16 * 10^-2 V

For combustion of Mg:

Mg (s) + $\frac{1}{2}{O}_2\left(g\right)\to MgO\left(s\right)$

Here,   $\Delta ng=-\frac{1}{2}$

Now using

$\Delta H=\Delta U+\Delta ngRT$

-601.7 =   $\Delta U+\left(-\frac{1}{2}\right)*8.3*10^{-3}*300$

$\Delta U=-600.45kJ$

So; magnitude of   $\Delta U$ is 600 kJ (the nearest integer).