Solutions

  $\Delta T_f=k_f.m$

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

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

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

Depression in freezing point will be maximum for Al₂ (SO₄)₃ since its Van't Hoff factor is the highest. So its solution will have the lowest freezing point.

Hence, electronic configuration of CO²⁺ is $\left[Ar\right]3d^{7}4s^0$ . In complex ${\left[Co{\left(CN\right)}_6\right]}^{4-},$ given ligand CN^- is $\therefore$ strong hence, after pairing in d-subshell, total number of unpaired electron =

spin magnetic moment = $\sqrt[]{1\left(1+2\right)}=\sqrt[]{3}=1.73BM$ $=1.73BM\approx 2.0BM$  

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

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

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

KCl solution has molality (m) = 3.3, [Means 3.3 mol of KCl dissolved in 1 kg of solved]

Total mass of solution = mass of solute + mass of solvent

= 3.3 * 74.5 + 1000 gm

= 1245.85 gm

Volume of solution = $\frac{Mass}{density}=\frac{1245.85}{1.2}=1038.20ml$

$M=\frac{moles*1000}{V\left(ml\right)}=\frac{3.3*1000}{1038.2}=3.17M$          

Ans. = 3 (the nearest integer)

All the solution have higher ion production with respect to 0.1 M C₂H₅OH i.e. 0.1 M Ba₃ (PO₄)₂, 0.1 M Na₂SO₄, 0.1 M KCl and 0.1 M Li₃PO₄. Hence all have lowered freezing point than 0.1 M C₂H₅OH (which is non-ionisable in aqueous medium).

Ans. = 4

$\Delta T_f=0.2°C$

$m=\frac{0.7\times 1000}{93\times 42}=\frac{700}{93\times 42}=0.179m$

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

$i=\frac{\Delta T_f}{K_f\times m}=\frac{0.2}{1.86\times 0.179}=0.6$

$\frac{\alpha }{2}=1-0.6=0.4$

$\therefore \alpha =0.4\times 2=0.8$

% of = 80%

  $\Delta T_b=i\times K_b\times m$

 or $\Delta T_b=K_b\times m$   (i = 1 for non-electrolyte)

  $or\frac{{\left(\Delta T_b\right)}_A}{{\left(\Delta T_b\right)}_B}=\frac{{\left(K_b\right)}_A}{{\left(K_b\right)}_B}$

  $or\frac{x}{y}=\frac{1}{8}\left|\therefore y=8\right|$              

Molality itself a strength representing terms for solution which does not depend upon the temperature.

Let we take $$ of solution

Mass of solute = Volume × Density

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

= 0.525 gram

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

Mass of solvent = 1000 – 0.525 = 999.475 gram

$\Rightarrow i=1.9$