Chemistry Solutions

Moles of SO₂ = $\frac{224*10^{-3}}{22.4}$

=0.01 mole

Moles of NaOH = 0.1 * 0.1

= 0.01 mole

SO₂+NaOHNaHSO₃

0.01 mole0.01 mole-

-0.01 mole

Non-volatile solute is NaHSO3

Moles of water = $\frac{36}{18}=2$

Using ; relative lowering in V.P

$\frac{P^o-P}{P^o}=i{x}_B$

Where; $\Delta P=P^0-P$ is lowering in V.P

$\Delta P=P^{0}i{x}_B$

i for NaHSO₃ = 2

here; $x_B=\frac{n_B}{n_A}$ since solution is dilute

$\Delta P=24*2*\frac{0.01}{2}=0.24$

$\Delta P=24*10^{-2}mmHg$

So; x = 24

$\Delta T_f=0.5°C$

K_f = 1.86

Using, density of water = 1g / mL

i = ?

  $\Delta T_f=i{k}_f.m$           

  $0.5=i*1.86*\frac{9.45}{94.5*500}*1000$         

i = 1.344

Now, using $\alpha =\frac{i-1}{n-1}$  

n for ClCH₂COOH = 2

a =  $\frac{1.344-1}{2-1}=0.344$  

Using

$K_a=\frac{C{\alpha }^2}{1-\alpha }$  

$K_a=\frac{0.2*{\left(0.344\right)}^3}{0.66}=36*10^{-3}$       

So; x = 36

$\Delta T_f=K_f*m*i$

0.93 = 1.86 * 1 * i

i = $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ $i=1+\left(\frac{1}{n}-1\right)$ $\raisebox{1ex}{$$}\!\left/ \!\raisebox{-1ex}{$$}\right.$

$\therefore n=2$

  $\Delta T_f=k_f.m$

$T_f^0-T_f=5.12*\frac{\left(\frac{10}{58}\right)}{\left(\frac{200}{1000}\right)}$

    5.5 – T_f = $\frac{5.12*5*10}{58}$  

  $T_f=1.086°C=\left(1.086\right)°C\cong 1°C$             

$AB?A^{+}+B^{-}$

1-x x x   $\therefore i=1+x$

$\begin{array}{l}\Delta T_b=i*k_b*m\\ 2.5=\left(1+x\right)*0.52m\end{array}$

$\therefore molality=\frac{2.5}{1.75*0.52}=2.74$

the nearest integer is 3.

$A_2{B}_3?2A^{+3}+3B^{-2}$

1-α        2α        3α

$\therefore i=1+4\alpha =1+4*0.6=1+2.4=3.4$

$\Delta T_b=i{k}_{b}m=3.4*0.52*1=1.768\approx 1.77K$

$T_b=374.77K\approx 375K.$         

Let we take  $1\mathcal{l}$  of solution

Mass of solute = Volume * Density

= 0.5  $m\mathcal{l}*1.05gm/m\mathcal{l}$  

= 0.525 gram

Mass of solution = 1 kg. [considering very dilute solution]

Mass of solvent = 1000 – 0.525 = 999.475 gram

$\Rightarrow i=1.9$  

$m=\frac{w*1000}{molecularwt*w_{solvent}}=\frac{10.2*1000}{176*150}$  

$=\frac{1020\cancel{0}}{176*15\cancel{0}}=0.386$  

  $\Delta T_f=K_f*m$

3.9 * 0.386

$\therefore x*10^{-1}=15.05*10^{-1}$  

Ans = 15

Total pressure of mixture =   $P_{Total}=X_A*P_A^o+X_B*P_B^o=90*0.6+15*0.4=60mm$

Mole fraction of B in vapour phase, $\left(X_B^{\text{'}}\right)=\frac{X_B*P_B^o}{P_{Total}}$

$\therefore X_B^{\text{'}}=\frac{0.4*15}{60}=0.1=1*10^{-1}$  

Peptide contains four amino acid i.e. glycine, aspartic acid and histidine, so it will have three  peptide linkage.