Chemistry Solutions

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.

$Molality=\frac{molesofsolute*1000}{wtofsolvent\left(gm\right)}$  

$Molality=\frac{35*1000*1.46}{36.5*100}$ = 14.0 M

$\Delta T_b=i{K}_{b}m$

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

$\therefore \frac{4}{4}=\frac{K_b*1.5}{K_f*4.5}$

$\frac{K_b}{K_f}=3$

$i=\frac{m_{theor}}{m_{app}}=\frac{30}{m_{app}}$

$\Delta T_b=i{k}_b.m]\underset{m-x}{\overset{m}{HA}}=\underset{x}{\overset{-}{H^{+}}}+\underset{x}{\overset{-}{A^{-}}}$

$0.0156=\left(\frac{m*x}{m}\right)*0.52*m$

m + x = 0.03

$\Rightarrow x=0.01\to i=1.5=\frac{30}{m_{app}}\Rightarrow m_{appp}=20$

At pH = 6.4

As in case of H₂CO₃,

pH = pKa it will be only when

[weak acid] = [conjugate base].

In case of 2

H₂PO^-₄ |HPO^2-₄

Solute and solvent both should obey Raoult's law

$\to$ ${\pi }_{{\mathrm{K}\mathrm{N}\mathrm{O}}_3}={\pi }_{\text{glucose.}}$

$(\mathrm{i}\mathrm{C}\mathrm{R}\mathrm{T}{)}_{{\mathrm{K}\mathrm{N}\mathrm{O}}_3}=(\mathrm{C}\mathrm{R}\mathrm{T}{)}_{\text{glucose}}$

 $\frac{}{}\frac{}{}$ $\frac{\mathrm{i}*1.34}{101.1}*\mathrm{R}\mathrm{T}=\frac{4.77}{180}*\mathrm{R}*\mathrm{T}$

$\mathrm{i}=\frac{4.77}{180}*\frac{101.1}{1.34}=2$

Hence, extent of dissociation is $100\mathrm{\%}$

∴ % of S in the compound
= (32/233) * (mass of BaSO? / mass of compound) * 100 = (32 * 0.35 * 100) / (233 * 0.25) = 19.227 ≈ 19.23