Chemistry Solutions

  $\Delta T_b=i{K}_{b}m$ ${}_{}{}_{}$

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

For acetone solution,

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

For Benzene solution,

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

$\Delta T_b=i*K_b*m$

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

= $0.13°C=13*10^{-2}°C$ ${}^{}$

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

$\therefore x=13$

Simple distillation can be applied for the separation of two liquids has boiling point difference greater than 20°C.

Boiling point of propanol > boiling point of propane (due to H-bonding effect)

Molality (m) = $\frac{\left(40/180\right)}{0.2}=\left(\frac{10}{9}\right)molal$

$\Delta T_f=T_f-T_f^{\text{'}}=1.86*\frac{10}{9}K$

$T_f^{\text{'}}=273.15-1.86*\frac{10}{9}K$

$=271.08K\approx 271K$

Let mass of water initially = x gram

Mass of sucrose = (1000 – x) gram

Mole of sucrose = $\frac{1000-x}{342}mol$

$?$ 0.75 molal Þ 0.75 mole solute in 1 kg of solvent

$0.75=\frac{\left(1000-x\right)/342}{x/1000}\left[molality=\frac{n_{solute}}{M_{solvent}\left(kg\right)}\right]$

$4=\frac{\frac{\left(1000-795.86\right)*1.86}{342}}{a}$        

a = 0.2775kg or 277.5 gram

Ice separated = (795.86 – 277.5) gram

= 518.3 gram

Ans. = 518 (the nearest integer)

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

  $\Delta T_f=i*K_f*m$

{for ethylene glycol I = 1, due to non-ionisable}

$=1*1.86*\frac{83*1000}{62*625}K$     

$=3.984K\approx 4K$ (the nearest integer)

Hence, freezing point of solution = 273 K – 4K = 269 K.

$\Delta T_f=0.2°C$

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

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

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

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

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

% of = 80%

$\Delta T_f=i*k_f*M$

$\frac{{\left(\Delta T_f\right)}_x}{{\left(\Delta T_f\right)}_y}=\frac{i{K}_f{m}_x}{i{K}_f{M}_Y}=\frac{M_X}{M_Y}$

$\Rightarrow \frac{1}{4}=\frac{1}{M_X}*\frac{M_Y}{1}\Rightarrow \frac{M_X}{M_Y}=\frac{4}{1}=\frac{1}{0.25}$