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The azeotropic mixture of water (B.P. = 100^oC) and HCl (B.P. = 86^oC) boils at about 120^oC. During fractional distillation of this mixture it is possible to obtain:
1. pure HCl
2. pure H₂O
3. pure H₂O as well as pure HCl
4. neither H₂O or HCl

Azeotropic mixture of water and HCl boils at 381.5 K. By distillation the mixture it is possible to obtain :
1. Pure HCl only
2. Pure water only
3. Neither HCl nor water
4. Both water and HCl in pure state

When a liquid that is immiscible with water was steam distilled at 95.2^oC at a total pressure of 748 torr, the distillate contained 1.25 g of the liquid per gram of water . The vapour pressure of water is 648 torr at 95.2^oC, what is the molar mass of liquid ?
1. 7.975 g/mol
2. 166 g/mol
3. 145.8 g/mol
4. None of these
 

If two liquids A $\left({\mathrm{P}}_{\mathrm{A}}^0=100 \mathrm{torr}\right)$ and B $\left({\mathrm{P}}_B^0=200 \mathrm{torr}\right)$ are completely immiscible with each other ( each one will behave independtly of the other ) are present in a closed vessel. The total vapour pressure of the system will be :
1. less than 100 torr
2. greater than 200 torr
3. between 100 to 200 torr
4. 300 torr

Water and chlorobenzene are immiscible liquids. Their mixture boils at 89^oC under a reduced pressure of $7.7\times {10}^4 \mathrm{Pa}$. The vapour pressure of pure water at 89^oC is $7\times {10}^4 \mathrm{Pa}$. Weight percent of chlorobenzene in the distillate is :
1. 50
2. 60
3. 78.3
4. 38.46

The degree of dissociation of an electrolyte is $\mathrm{\alpha }$ and its van't Hoff factor is i. The number of ions obtained by complete dissociation of 1 molecule of the electrolyte is :
1. $\frac{\mathrm{i}+\mathrm{\alpha }-1}{\mathrm{\alpha }}$
2. $\mathrm{i}-\mathrm{\alpha }-1$
3. $\frac{\mathrm{i}-1}{\mathrm{\alpha }}$
4. $\frac{\mathrm{i}+1+\mathrm{\alpha }}{1-\mathrm{\alpha }}$

One mole of a solute A is dissolved in a given volume of a solvent. The association of the solute take place as follows :
$\mathrm{nA}\rightleftharpoons {\mathrm{A}}_{\mathrm{n}}$
If $\mathrm{\alpha }$ is the degree of association of A, the van't Hoff factor i is expressed as :
1. $\mathrm{i}=1-\mathrm{\alpha }$
2. $\mathrm{i}=1+\frac{\mathrm{\alpha }}{\mathrm{n}}$
3. $\mathrm{i}=\frac{1-\mathrm{\alpha }+{\displaystyle \frac{\mathrm{\alpha }}{\mathrm{n}}}}{1}$
4. i = 1

The van't Hoff factor i for an electrolyte which undergoes dissociation and association in solvent are respectively :
1. greater than one and less than one
2. less than one and greater than one
3. less than one and less than one
4. greater than one and greater than one

Bromoform has a normal freezing point of 7.734^oC/m and it's Kf = 14.4^oC m. A solution of 2.60 g of an unknown 100 g of bromoform freezes at 5.43^oC. What is the molecular weight of the unknown?
1. 16.25
2. 162.5
3. 100
4. None of these

${\mathrm{C}}_6{\mathrm{H}}_6$ freezes at 5.5$°$C. At what temperature will a solution of 10.44 g of ${\mathrm{C}}_4{\mathrm{H}}_{10}$ in 200 g of ${\mathrm{C}}_6{\mathrm{H}}_6$ freeze ? K_f (${\mathrm{C}}_6{\mathrm{H}}_6$) = 5.12$°$C/m
1. 4.60$°$C
2. 0.89$°$C
3. 5.50$°$C
4. None of the above

How much ethyl alcohol must be added to 1.0 L of water so that solution will not freeze at -4$°$F ? (K_f = 1.86$°$C/m)
1. < 20 g
2. < 10.75 g
3. < 494.5 g
4. > 494.5 g

When 36.0 g of a solute having the empirical formula ${\mathrm{CH}}_2\mathrm{O}$ is dissolved in 1.20 kg of water, the solution freezes at -0.93$°$C. What is the molecular formula of the solute ?(K_f  =1.86$°$C kg mol^-1)
1.${\mathrm{C}}_2{\mathrm{H}}_4\mathrm{O}$
2. ${\mathrm{C}}_2{\mathrm{H}}_2{\mathrm{O}}_2$
3. ${\mathrm{C}}_2{\mathrm{H}}_4{\mathrm{O}}_3$
4. ${\mathrm{C}}_2{\mathrm{H}}_4{\mathrm{O}}_2$

A complex if represented as ${\mathrm{CoCl}}_3.{\mathrm{xNH}}_3$. Its 0.1 molal solution in water shows $∆{\mathrm{T}}_{\mathrm{f}}=0.558 \mathrm{K}$ K_f for ${\mathrm{H}}_2\mathrm{O}$ is 1.86 K molality^-1. Assuming 100 % ionisation of complex and co-ordination number of Co is six, calculate formula of complex :
1. $\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_6\right]{\mathrm{Cl}}_3$
2. $\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_5\mathrm{Cl}\right]{\mathrm{Cl}}_2$
3. $\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_4{\mathrm{Cl}}_2\right]\mathrm{Cl}$
4. None of these 
$\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_5\mathrm{Cl}\right]{\mathrm{Cl}}_2$
 
 

The freezing point of equimolal aqueous solutions will be highest for :
1. ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}_3\mathrm{Cl}$
2. $\mathrm{Ca}{\left({\mathrm{NO}}_3\right)}_2$
3. $\mathrm{La}{\left({\mathrm{NO}}_3\right)}_2$
4. ${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6$

The freezing point of a 4% aqueous solutions of 'A' is equal to the freezing point of 10% aqueous solution of 'B'. If the molecular weight of 'A' is 60, then the molecular weight of 'B' will be:
1. 160
2. 90
3. 45
4. 180

Depression in freezing point of 0.01 molal aqueous HCOOH solution is 0.02046. 1 molal aqueous urea solution freezes at -1.86^oC, assuming molality equal to molarity, pH of HCOOH solution is :
1. 2
2. 3
3. 4
4. 5

When mercuric iodide is added to the aqueous solution of KI, then the :
1. freezing point is raised
2. freezing point is lowered
3. freezing point does not change
4. boiling point does not change 

Dimer of acetic acid in liquid benzene is in equilibrium with acetic acid monomer at certain temperature and pressure. If 25% of the dimer molecules are separated out then
1. Freezing point of the solution reduces
2. Average molar mass of solute increases
3. Boiling point of solution increases
4. Molar mass of solute decreases

The solubility of a specific non-volatile salt is 4 g in 100 g of water at 25$°$C. If 2.0  g , 4.0 g and 6.0 g of the salt added of 100 g of water at 25$°$C, in system X,Y and Z. The vapour pressure would be in the oreder :
1. X < Y < Z
2. X > Y > Z
3. Z > X = Y
4. X > Y = Z

6.0 g of urea (molecular weight = 60) was dissolved in 9.9 moles of water. If the vapour pressure of pure water is P$°$, the vapour pressure of solution is 
1. 0.10 P$°$
2. 1.10 P$°$
3. 0.90 P$°$
4. 0.99 P$°$

The vapour pressure of an aqueous solution of sucrose at 373 K is found to be 750 mm Hg.
The molality of the solution at the same temperature will be:

Estimate the lowering of vapour pressure due to the solute (glucose) in a 1.0 M aqueous solution at 100^oC:
1. 10 torr
2. 18 torr
3. 13.45 torr
4. 24 torr

Calculate the weight of non-volatile solute having molecular weight 40, which should be dissolved in 57 gm octane to reduce its vapour pressure to 80 %
1. 47.2 g
2. 5 g
3. 106.2 g
4. None of these

Equal weight of a solute are dissolved in equal weight of two solvents A and B are formed very dilute solution. The relative lowering of vapour pressure for the solution B has twice the relative lowering of vapour pressure for the solution A. If M_A and M_B are the molecular weights of solvents A and B respectively, then
1. ${\mathrm{M}}_{\mathrm{A}}={\mathrm{M}}_{\mathrm{B}}$
2. ${\mathrm{M}}_{\mathrm{B}}=2\times {\mathrm{M}}_{\mathrm{A}}$
3. ${\mathrm{M}}_{\mathrm{A}}=4{\mathrm{M}}_{\mathrm{B}}$
4. ${\mathrm{M}}_{\mathrm{A}}=2{\mathrm{M}}_{\mathrm{B}}$

An ideal solution has two components A and B. A is more volatile than B, i.e. ${\mathrm{P}}_{\mathrm{A}}^{\mathrm{o}}>{\mathrm{P}}_{\mathrm{B}}^{\mathrm{o}}$ and also ${\mathrm{P}}_{\mathrm{A}}^{\mathrm{o}}>{\mathrm{P}}_{\mathrm{total}}$. If XA and YA are mole fractions of components A in liquid and vapour phases, then :
1. ${\mathrm{X}}_{\mathrm{A}}={\mathrm{Y}}_{\mathrm{A}}$
2. ${\mathrm{X}}_{\mathrm{A}}>{\mathrm{Y}}_{\mathrm{A}}$
3. ${\mathrm{X}}_{\mathrm{A}}<{\mathrm{Y}}_{\mathrm{A}}$
4. Data insufficient

At 28oC, the vapour pressure of pure liquid A (mol.wt = 40) is 100 torr, while that of pure liquid B is 40 torr, (mol. wt = 80). The vapour pressure at 25^oC of a solution containing 20 g of each A and B is:
1. 80 torr
2. 59.8 torr
3. 68 torr
4. 48 torr