Colligative Properties and Molar Mass

0.5 % KCl solution has molality (m) = $\frac{0.5*1000}{74.5*99.5}$  

$KC{l}_{\left(aq\right)}?K_{\left(ag\right)}^{+}+C{l}_{\left(aq\right)}^{-}$  

1 - a        a          a

And I =  $\left(1-\cancel{\alpha }+\cancel{\alpha }+\alpha \right)=1+\alpha$  

$i=\frac{\Delta Tf}{kf*m}=\frac{0.24*74.5*99.5}{1.8*0.5*1000}=1+\alpha$  

1.976 = 1 + a

$\therefore \alpha =0.976$  

% = 97.6%

the nearest 98.

$\Delta T_f=0.5°C$

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

Kindly consider the following Image 

$\Delta T_f=K_f*m*i$

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.$

$\therefore n=2$

  $\Delta T_b=i{K}_{b}m$ ${}_{}{}_{}$

$m=\frac{w*1000}{M*W_{solvent}}$

For acetone solution,

$0.17=1*1.7*\frac{1.22*1000}{M*100}$

For Benzene solution,

$2Acid\to {\left(Acid\right)}_2\therefore i=1/2$

$\Delta T_b=i*K_b*m$

$=\frac{1}{2}*2.6*\frac{1.22*1000}{122*100}°C$

= $0.13°C=13*10^{-2}°C$ ${}^{}$

$\therefore x*10^{-2}=13*10^{-2}$

$\therefore x=13$

$\Delta T_f=i{K}_{f}m$

Higher the value of i, more be $\Delta T_f.$

Glucose →i = 1

Hydrazine → I = 1

Glycine → I = 1

 KHSO₄ → I = 3 $\left(K^{+}{H}^{+}S{O}_4^{--}\right)$

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

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

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%