Solutions Test

Result

Solutions Test - 7

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0.25

This section contains multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct

The vapour pressure of pure liquid solvent is 0.80 atm. When a non-volatile solute B is added to the solvent, its vapour pressure drops to 0.60 atm. Thus, mole fraction of the component B is

A
0.50

B
0.25

C
0.45

D
0.75

Solution

Using Raoults law for lowering of vapour pressure.We get
putting,Values of actual and lowered vapour pressure we get

where x₂ is the mole fraction .
Question 2
1 / -0.25

Vapour pressure of pure water at 298 K is 23.8 torr. Thus, vapour pressure of a solution containing 6 g urea and 68.4 g sucrose in 10 moles of water is

A
23.0 torr

B
19.2 torr

C
21.0 torr

D
23.1 torr

Solution

By Raoult ’s law, for a mixture of non-volatile solute in volatile solvent.
Question 3
1 / -0.25

Vapour pressure of nitrobenzene (molar mass = 123 g mol^-1 ) is 3.6 kNm^-2 and that of water (molar mass = 18 g mol^-1 ) is 977 kNm^-2 . They form immiscible mixture at the given temperature. Thus, percentage of nitrobenzene in the vapour phase is

A
20.12%

B
40.28%

C
79.88%

D
59.72%

Solution

For immiscible solvents, composition of each component is based on
Question 4
1 / -0.25

The vapour pressure of n-hexane at 350 K is 840 torr and that of cyclohexane is 600 torr. Mole fraction of hexane in the mixture that boils at 350 K and 1 atm pressure assuming ideal behaviour is

A
0.75

B
0 .50

C
0 .80

D
0 .67

Solution

d) Mixture boils at 1 atm and 350 K.
Thus, total vapour pressure is 1 atm = 760torr (h) used for hexane and (ch) for cyclohexane.
Question 5
1 / -0.25

At 298 K, the vapour pressure of pure water is 23.76 mm Hg and that of an aqueous solution is 22.98 mm Hg. Thus, molality of solution is (assume dilute solution)

A
0.033

B
0.67

C
1.83

D
5.55

Solution

(c)
By Raoult ’s law for a non-volatile solute in volatile solvent.
Question 6
1 / -0.25

A binary liquid solution is prepared by mixing n-heptane and ethanol. Which one of the following statements is/are correct regarding the behaviour of the solution?

A
The solution formed is an ideal solution

B
The solution is non ideal showing positive deviation from raults law

C
The solution is non ideal showing negative deviation from raults law

D
n - heptane shows positive wheres ethanol shows negative deviation from raults law

Solution

Ethanol molecules are joined to each other by intermolecular H-bonding thus, they have high boiling point. When n-hexane is mixed into it, H-bonding breaks. Since, force of attraction between n-hexane and ethanol molecules is smaller than between the hexane-hexane and ethanol-ethanol. Thus, mixture is more volatile. This shows positive deviation from Raoult ’s law
Question 7
1 / -0.25

18 g of glucose (C₆ H₁₂ O₆ ) is added to 178.2 g of water. The vapour pressure of water for this aqueous solution at 100 °C is

A
759.00 torr

B
7.60 torr

C
76.00 torr

D
752.40 torr

Solution

(d)
At 100 °C, vapour pressure of water = 1 atm. Thus, p⁰ = 760torr Let vapour pressure of the solution = p_s
Question 8
1 / -0.25

Which of the following liquid pairs shows a positive deviation from Raoult ’s law?

A
Water+HCl

B
Benzene+methanol

C
Water+nitriic acid

D
Acetone+chloroform

Solution

(a) When HCI is mixed with H₂ 0, force of attraction between HCI and H₂ O molecules is increased (due to dissolution), hence vaporisation decreased, a case of negative deviation.
(b) CH₃ OH molecules are joined by intermolecular H-bonding which provides it a liquid state. On adding benzene, H-bonding breaks and force of attraction between CH₃ OH and benzene molecules is less than between CH₃ OH-CH₃ OH and between benzene-benzene hence, more vaporisation of mixture, a case of positive deviation.
(c) As in (a), negative deviation.
(d) CH₃ COCH₃ and CHCI₃ are volatile. On mixing, there is increase in intermolecular H-bonding due to the formation of chloretone (A)
Thus, force of attraction between (CHCI₃ -CH₃ OCH₃ ) molecules is larger than between (CHCI₃ -CHCI₃ )and (CH₃ COCH₃ - CH₃ OC H₃ ) molecules. Thus, vaporisation decreases, a case of negative deviation.
Question 9
1 / -0.25

A solution of two liquids boils at a temperature more than the boiling point of either of them. Hence, the binary solution shows

A
Negative deviation from Raoult ’s law

B
Positive deviation from Raoult ’s law

C
No deviation from Raoult ’s law

D
Positive or negative deviation depending on composition.

Solution

If force of attraction between A and B (solvents) is higher than that between A and A or B and B, then there is decrease in vaporisation.
Thus, higher temperature (i.e. boiling point) is required to convert liquids into vapour state. This is a case of negative deviation from Raoult 's law
Question 10
1 / -0.25

Lowering of vapour pressure of an aqueous solution of a non-volatile non-electrolyte 1 molal aqueous solution at 100 °C is

A
14.12 torr

B
312 torr

C
13.44 torr

D
352 torr

Solution

(c) Molality = 1 mol kg^-1 (solvent) Thus, solute = 1 mole