Solutions Test

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Solutions Test - 2

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following plots represents an ideal binary mixture?

A
Plot of P_total v/s 1/X_B is linear (X_B = mole fraction of 'B' in liquid phase).

B
Plot of P_total v/s Y_A is linear (Y_B = mole fraction of 'A' in vapour phase)

C
Plot of 1/P_total v/s Y_A is linear

D
Plot of 1/P_total v/s Y_B is non-linear

Solution

By raoult’s law:

P_total = P °_A X_A + P °_B X_B

Y_A = P °_A X_A / P_total

As it is ideal solution. The value of X_A vary linearly.

Hence plot of 1/P_total v/s Y_A is linear
Question 2
1 / -0

How many mL of 2.50% (m/v) salt solution would contain 1.80 g of salt?

A
105 mL

B
104 mL

C
45.0 mL

D
72.0 mL

Solution

% m/v = (mass of solute (g)) / volume of solution (ml) X 100

as mass of solute = 1.80 g

% m/v = 1.8/volume × 100 = 2.5

therefore, volume = 1.8 × 100 / 2.5 = 72 ml
Question 3
1 / -0

If mole fraction of NaCl_(aq) is same as that of water then molality of NaCl is

A
5.55

B
55.55

C
0.18

D
58.5

Solution

As mole fraction NaCl = mole fraction water

Also sum of mole fractions = 1

Therefore, mole fraction of water = mole fraction of NaCl = 0.5

Let both of them has 1 mol each in solution.

Moles of NaCl = 1 mol = moles of water.

At standard temperature and pressure volume of 1 mol water is 18 ml = 0.018 L

Molarity = moles of NaCl / volume of solution = 1 / 0.018 = 55.55 M
Question 4
1 / -0

An ideal mixture of liquids A and B with 2 moles of A and 2 moles of B has a total vapour pressure of 1 atm at a certain temperature. Another mixture with 1 mole of A and 3 moles of B has a vapour pressure greater than 1 atm. But if 4 moles of C are added to the second mixture, the vapour pressure comes down to 1 atm. Vapour pressure of C in pure state is 0.8 atm. Calculate the vapour pressure of pure A and pure B: -

A
_P^o =1.4atm, _P^o=0.7atm

B
_P^o=1.2atm, _P^o=0.6atm

C
_P^o =1.4atm, _P^o=0.6atm

D
_P^o=0.6atm, _P^o=1.4atm

Solution

In solution I: X_A = 1/2, X_B = 1/2.

P °_A be vapour pressure of pure A.

P °_B be vapour pressure of pure B.

P °_C be vapour pressure of pure C.

By Raoult’s law P_total = P °_A X _A + P °_B X_B

1 = ( P °_A + P °_B )/2

2 = P °_A + P °_B ……… (I)

In solution II:

Total moles = 1 + 3 + 4 = 8 moles

X_A = 1/8, X_B = 3/8, X_C = 4/8.

By Raoult’s law, P_total = P °_A X_A + P °_B X_B + P °_C X_C

1 = P °_A x 1/8 + P °_B x 3/8 + P °_C x 4/8

8 = P °_A + 3P °_B + 0.8x4

P °_A + 3P °_B = 4.8………… (II)

From equations I and II.

P °_A = 0.6atm

P °_B = 1.4atm

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Solutions Test - 2

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Solutions Test - 2

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Question 1

Plot of Ptotal v/s 1/XB is linear (XB = mole fraction of 'B' in liquid phase).

Plot of Ptotal v/s YA is linear (YB = mole fraction of 'A' in vapour phase)

Plot of 1/Ptotal v/s YA is linear

Plot of 1/Ptotal v/s YB is non-linear

Solution

Question 2

105 mL

104 mL

45.0 mL

72.0 mL

Solution

Question 3

5.55

55.55

0.18

58.5

Solution

Question 4

Po =1.4atm, Po=0.7atm

Po=1.2atm, Po=0.6atm

Po =1.4atm, Po=0.6atm

Po=0.6atm, Po=1.4atm

Solution

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Plot of P_total v/s 1/X_B is linear (X_B = mole fraction of 'B' in liquid phase).

Plot of P_total v/s Y_A is linear (Y_B = mole fraction of 'A' in vapour phase)

Plot of 1/P_total v/s Y_A is linear

Plot of 1/P_total v/s Y_B is non-linear

_P^o =1.4atm, _P^o=0.7atm

_P^o=1.2atm, _P^o=0.6atm

_P^o =1.4atm, _P^o=0.6atm

_P^o=0.6atm, _P^o=1.4atm