Solutions

Let volume of solution = x ml
So mass of solution = 1.2x
And mass of water = x gm
Mass of solute = 0.2x
Molality = (W_solute * 1000) / (M_solute * W_solvent) = (0.2x * 1000) / (40 * x) = 5 m

? = CRT
2.42 * 10? ³ = ( (1.46/M_polymer) / 0.1 ) * 0.083 * 300
M_polymer = (1.46 * 0.083 * 300) / (2.42 * 10? ³ * 0.1)
= 14.96 * 10? gm = 15 * 10? gm

ΔTb = 0.6 K
ΔTb = Kb * m = Kb * (wt. * 1000)/ (mol wt. * Wsolvent (gm)
0.6 = 5 * (3 * 1000)/ (mol wt. * 100)
∴ mol wt. = (15 * 10)/ (0.6) = 1500/6
= 250 g/mole

  $\Delta T_f=k_f.m$

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

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

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

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

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

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

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

$\Delta T_f=0.2°C$

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

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

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

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

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

% of = 80%