Chemistry Solutions

Let we take  $1\mathcal{l}$  of solution

Mass of solute = Volume * Density

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

= 0.525 gram

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

Mass of solvent = 1000 – 0.525 = 999.475 gram

$\Rightarrow i=1.9$  

$m=\frac{w*1000}{molecularwt*w_{solvent}}=\frac{10.2*1000}{176*150}$  

$=\frac{1020\cancel{0}}{176*15\cancel{0}}=0.386$  

  $\Delta T_f=K_f*m$

3.9 * 0.386

$\therefore x*10^{-1}=15.05*10^{-1}$  

Ans = 15

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, $\left(X_B^{\text{'}}\right)=\frac{X_B*P_B^o}{P_{Total}}$

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

Peptide contains four amino acid i.e. glycine, aspartic acid and histidine, so it will have three  peptide linkage.

  $\Delta T_b=i{K}_{b}m$ ${}_{}{}_{}$

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

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 $\left(K^{+}{H}^{+}S{O}_4^{--}\right)$

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)