Solutions Test

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Solutions Test - 2

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Weekly Quiz Competition

Question 1
1 / -0

If 510 mg of a liquid on vapourisation in Victor Meyer's apparatus displaces 67.2 cm³ of air at STP, what is the molecular weight of the liquid?

A
1700

B
170

C
17

D
130

Solution
Question 2
1 / -0

Directions: The following question has four choices out of which ONLY ONE is correct.

Which one of the following aqueous solutions will exhibit the highest boiling point?

A
0.01 M Na₂SO₄

B
0.01 M KNO₃

C
0.015 M urea

D
0.015 M glucose

E
0.0015 M glucose

Solution
Question 3
1 / -0

van 't Hoff factor of BaCl₂ at 0.01 M concentration is 1.98. What is the percentage of dissociation of BaCl₂ at this concentration?

A
49

B
89

C
98

D
100

Solution

Question 4

1 / -0

Directions: Match the statements/expressions in Column – I with the statements/expressions in Column – II and indicate your answer by selecting the appropriate option. Any statement in Column – I can correctly match with one or more statement in Column – II.

Column – I	Column – II
(A) Ideal solution	(a) Raoult's law
(B) Non-ideal solution	(b) ΔH ≠ 0
(C) Degree of dissociation of solute	(c) Solutions with the same vapour pressure at a temperature
(D) Isopiestic solution	(d) ΔH = 0
	(e) Van't Hoff factor

(A) – (a), (d); (B) – (b); (C) – (e); (D) – (c)

(A) – (b), (c); (B) – (d), (c); (C) – (b); (D) – (a)

(A) – (c), (a); (B) – (b), (d); (C) – (a); (D) – (b)

(A) – (d), (b); (B) – (c), (e); (C) – (d); (D) – (d)

Solution

(A) An ideal solution obeys Raoult's law (a) and H = 0 for the formation of such solutions (d).
Thus, (A) – (a), (d)
(B) Non-ideal solutions do not obey Raoult's law and for such solutions, H 0 during their formation (b).
Thus, (B) – (b)
(C) Degree of dissociation of an electrolytic solute can be determined by van't Hoff factor (e).
Thus, (C) – (e)
(D) Solutions with the same vapour pressure at a particular temperature (c) are called isopiestic.
Thus, (D) – (c)

Question 5
1 / -0

Which one of the following pairs of solution can we expect to be isotonic at the same temperature?

A
0.1 M urea and 0.1 M NaCl

B
0.1 M urea and 0.2 M MgCl₂

C
0.1 M NaCl and 0.1 M Na₂SO₄

D
0.1 M Ca(NO₃)₂ and 0.1 M Na₂SO₄

Solution

'i' for Ca(NO₃)₂ and Na₂SO₄ is 3 each. Their colligative molarities are also the same. Hence, 0.1 M Ca(NO₃)₂ and 0.1 M Na₂SO₄ solutions are isotonic.
Question 6
1 / -0

Directions: The following question has four choices, out of which ONLY ONE is correct.

If a 6.84% (wt/vol.) solution of cane sugar (mol. wt = 342) is isotonic with 1.52% (wt/vol.) solution of thiocarbamide, then the molecular weight of thiocarbamide is

A
180

B
60

C
76

D
152

Solution
Question 7
1 / -0

Ethylene dibromide (C₂H₄Br₂) and 1,2-dibromopropane (C₃H₆Br₂) form a series of ideal solutions over the whole range of compositions and at 85^oC, the vapour pressures of these two pure liquids are 173 torr and 127 torr, respectively. If 10.0 g of ethylene bromide is dissolved in 80.0 g of 1,2-dibromopropane, then which of the following statements is correct?

A
Partial pressure of ethylene dibromide in vapour phase is 112 torr.

B
Partial pressure of 1,2-dibromopropane is 20.4 torr.

C
Total pressure of the vapour is 232.4 torr.

D
Mole fraction of ethylene dibromide in the vapour phase is 0.154.

Solution

= 0.154

Thus, option (4) is correct.

Question 8

1 / -0

Match the following:

Column – I	Column – II
(A) Positive deviation from Raoult's law	(a) Osmotic pressure method
(B) Addition of HgI₂ to KI solution	(b) Isotonic with human blood
(C) Accurate method of determination of molar mass of a protein	(c) Binary liquid mixtures
(D) 0.9% NaCl solution	(d) Elevation of freezing point of solution

(A) – (c), (B) – (d), (C) – (a), (D) – (b)

(A) – (b), (B) – (a), (C) – (d), (D) – (c)

(A) – (a), (B) – (b), (C) – (b), (D) – (d)

(A) – (d), (B) – (c), (C) – (c), (D) – (a)

Solution

(A) Binary liquid mixtures exerting a repulsive force on each other show positive deviation from Raoult's law as more molecules are sent to the vapour phase.
Thus, (A) – (c).
(B) Addition of HgI₂ to KI solution decreases the total concentration as
HgI₂(s) + 2 (K⁺ + I^-) 2K⁺ + [HgI₄]^2-.Thus, 3 ions are produced from 3 ions. So, the depression of freezing point becomes less, i.e. there will be elevation of freezing point of solution.
So, (B) – (d).
(C) Osmotic pressure method gives the most accurate molecular weight of proteins. So, (C) – (a).
(D) 0.9% NaCl or normal saline is isotonic with human blood. Thus, (D) – (b).

Question 9
1 / -0

Two solutions of a substance are mixed in the following manner: 480 mL of 1.5 M first solution + 520 mL of 1.2 M second solution. What is the molarity of the final mixture?

A
1.20 M

B
1.50 M

C
1.344 M

D
2.70 M

Solution
Question 10
1 / -0

Benzene and toluene form nearly ideal solutions. At 293 K, the vapour pressure of benzene is 75 torr and that of toluene is 22 torr. The partial vapour pressure of benzene at 293 K in a solution containing 78 g of benzene and 46 g of toluene (in torr) is

A
50

B
25

C
37.5

D
53.5

Solution

Molecular weight of benzene = 78 g
Molecular weight of toluene = 92 g
Mixture contains 78 g benzene = 1 mole benzene and 46 g toluene = 0.5 mole toluene
Total moles of benzene and toluene = 1.5 mol
Mole fraction of benzene in mixture (X_b) = =
Vapour pressure of pure benzene (P_b^o)= 75 torr
Partial vapour pressure of benzene = P_b^oX_b
= 75 x = 50 torr
Question 11
1 / -0

The elevation in boiling point of solution of 13.44 g of CuCl₂ in 1 kg of water is (K_b = 0.52 K kg/mol)

A
0.05 K

B
0.1 K

C
0.16 K

D
0.21 K

Solution

For CuCl₂ i = 3
= ik_bm = 3 x 0.52 x = 0.1558 ≈ 0.16 K
Question 12
1 / -0

A 35.4 ml HCl is required for the neutralisation of a solution containing 0.275 g of sodium hydroxide. The normality of hydrochloric acid is

A
0.97 N

B
0.142 N

C
0.194 N

D
0.244 N

Solution

Use the formula N₁V₁ = N₂V₂ to find the normality of HCl.
Given:
V₁ = volume of HCl = 35.4 ml
Mass of NaOH = 0.275g
Molecular mass of NaOH = 23 + 16 + 1 = 40
40 g of NaOH = 1 g-eq of NaOH
0.275 g of NaOH = eq
= m-eq
= 6.88 m-eq
N₁V₁ = N₂V₂
(HCl) = (NaOH)
Or N₁x 35.4 = 6.88 (m-eq = NV)
Or N₁ = = 0.194 N
Question 13
1 / -0

A solution containing 10g per dm³ urea (molecular mass = 60gmol^–1) is isotonic with a 5% solution of non volatile solute. The molecular mass of this non volatile solute is

A
300 g mol^–1

B
350 g mol^–1

C
200 g mol^–1

D
250 g mol^–1

Solution

Molar concentration of urea = per dm³
Molar concentration of volatile solute solution = per ml or per dm³
For solution of the same concentration or isotonic solution at the same temperature.
or M = 300 g mol^–1.
Question 14
1 / -0

An aqueous solution of glucose is 10% in strength. What will be the volume in which 1 g mole of it is dissolved?

A
9 litre

B
1.8 litre

C
8 litre

D
0.9 litre

Solution

10% glucose solution means 10 g glucose is dissolved in 100 ml solution (it is assumed that solution is dilute and density of solution is equal to density of water).
1 g mole glucose = 180 g glucose
180 g glucose is dissolved in 180 ml
= 1800 ml = 1.8 litre
Question 15
1 / -0

A 0.100 m solution of NaClO₃ freezes at - 0.343^oC. At 0.001 m concentration of the same salt, the electrical interferences between the ions no longer exist, because on an average, the ions are too far from each other.

The degree of dissociation of a 0.100 m NaCl solution is

A
42.5%

B
28.3%

C
85%

D
80%

Solution

m_effective = = = 0. 185 m
For NaClO₃, n = 2
i = 1 - + n = 1 + = = = 1.85
Hence, = 0.85, i.e. ^₌ 85%
Thus, (3) is the correct option.