# Liquid Solution

## The ratio of the value of any colligative property for K4 [Fe(CN)6] solution to that of Fe4[Fe(CN)6]3 (prussian blue), solution is nearly

Q: The ratio of the value of any colligative property for K4 [Fe(CN)6] solution to that of Fe4[Fe(CN)6]3 (prussian blue), solution is nearly

(A) 1

(B) 0.71

(C)1.4

(D) Less than 1

Sol:

Vont’s Hoff factor (i) for K4[Fe(CN)6] is 5 (assuming complete ionization)

K4[Fe(CN6]    →  4K+   +  [Fe(CN)6]-4

↓                          ↓

1 molecule       4 cations      1 anion

Similarly, i for Fe4[Fe(CN)6]3 is 7

As we know, colligative properties ∝ i

$\frac{i_{K_4[Fe(CN)_6}]}{i_{Fe_4[Fe(CN)_6}]_3} = \frac{Colligative \; Properties \; of \; K_4[Fe(CN)_6]}{Colligative \; Properties \; of \; Fe_4[Fe(CN)_6]_3} = \frac{5}{7} =0.71$

Ans: (B)

## A solution containing 28 g phosphorous in 315 g CS2 (b.p. = 46.3°C) boils at 47.9°C (Kb for CS2 is 2.34). What will be molecular formula of phosphorus?

Q: A solution containing 28 g phosphorous in 315 g CS2 (b.p. = 46.3°C) boils at 47.9°C (Kb for CS2 is 2.34). What will be molecular formula of phosphorus? (assuming complete association)

(A) P4

(B) P8

(C) P2

(D) None

Sol: Intermolecular forces among molecules of n-heptane and ethyl alcohols are weaker than among hexane and among ethyl alcohol molecules  ∴ V.P. of solution is more than V.P. of pure solvents.

Ans: (B)

## At 334 K the vapor pressure of benzene (C6H6) is 0.526 atm and that of toluene (C7H8) is 0.188 atm. In a solution containing 0.500 mole of benzene and 0.500 mole of toluene, what is the vapor pressure of toluene above the solution \at 334 K ?

Q: At 334 K the vapor pressure of benzene (C6H6) is 0.526 atm and that of toluene (C7H8) is 0.188 atm. In a solution containing 0.500 mole of benzene and 0.500 mole of toluene, what is the vapor pressure of toluene above the solution \at 334 K ?

(A) 0.188 atm

(B) 0.10 atm

(C) 0.357 atm

(D) 0.094

Sol: PT = P0T × XT

= 0.188 × 0.5

= 0.094

Ans: (D)

## At 25°C the vapor pressure of benzene, C6H6 (78g/mole), is 93.2 Torr and the of toluene, C7H8 (92 g/mol), is 28.2 Torr. A solution of 1.0 mole of C6H6 and 1.0 mol of C7H8 is prepared. Calculate the mole fraction of C6H6 in the vapor above this solution

Q: At 25°C the vapor pressure of benzene, C6H6 (78g/mole), is 93.2 Torr and the of toluene, C7H8 (92 g/mol), is 28.2 Torr. A solution of 1.0 mole of C6H6 and 1.0 mol of C7H8 is prepared. Calculate the mole fraction of C6H6 in the vapor above this solution (assume the solution is ideal).

(A) 0.607

(B) 0.768

(C) 0.232

(D) 0.393

Sol: $Y_B = \frac{P_B^o \times X_B}{P}$

$Y_B = \frac{93.2 \times 0.5}{P}$

$Y_T = \frac{28.2 \times 0.5}{P}$

$\frac{Y_B}{Y_T} = 3.3$

And, YB + YT = 1

On solving yB = 0.768

Ans: (A)

## Benzene (C6H6, 78 g/mol) and tolune (C7H8, 92 g/mol) form an ideal solution. At 60°C the vapor pressure of pure benzene and pure toluene are 0.507 atm and 0.184, respectively. The mole fraction of benzene in a solution of these two chemicals that has a vapor pressure of 0.350 atm at 60°C

Q: Benzene (C6H6, 78 g/mol) and tolune (C7H8, 92 g/mol) form an ideal solution. At 60°C the vapor pressure of pure benzene and pure toluene are 0.507 atm and 0.184, respectively. The mole fraction of benzene in a solution of these two chemicals that has a vapor pressure of 0.350 atm at 60°C

(A) 0.514

(B) 0.690

(C) 0.486

(D) 0.190

Sol: 0.35 = 0.507 × XB + 0.184 × Xt    ….(i)

XB + Xt = 1                   ….(ii)

On solving equation (1) and (2)

XB = 0.514

Ans:  (A)