Solutions Test

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Solutions Test - 6

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Weekly Quiz Competition

Question 1
4 / -1

Why is the molecular mass determined by measuring colligative property in case of some solutes is abnormal?

A
Due to association or dissociation of solute molecules

B
Due to insolubility of solute molecules

C
Due to decomposition of solute molecules

D
Due to large size of solute molecules

Solution

Due to association or dissociation of solute molecules there is a change in number of particles. Since colligative properties depend on number of particles there is a change in molecular mass.
Question 2
4 / -1

Which of the following representations of i (van't Hoff factor) is not correct?

A

B

C

D
Question 3
4 / -1

Which of the following relations is not correctly matched with the formula?

A
In case of association,

B
In case of dissociation,

C
Relative lowering of vapour pressure

D
Elevation in boiling point,

Solution

In case of dissociation: A_n → nA
Initial number of moles: 1 0
After dissociation: 1 − α nα
No. of particles = 1 − α + nα
Question 4
4 / -1

Which of the following will have same value of vant's Hoff factor as that of K₄[Fe(CN)₆]?

A
Al₂(SO₄)₃

B
AlCl₃

C
Al(NO₃)₃

D
Al(OH)₃

Solution

K₄[Fe(CN)₆] → 4K⁺ + [Fe(CN)₆]^4-
Al₂(SO₄)₃ → 2Al³⁺ + 3SO₄^2-
Question 5
4 / -1

Arrange the following aqueous solutions in the order of their increasing boiling points
(i) 10⁻⁴M NaCl
(ii) 10⁻⁴M Urea
(iii) 10⁻³M MgCl₂
(iv) 10⁻²M NaCl

A
(i) < (ii) < (iv) < (iii)

B
(ii) < (i) = (iii) < (iv)

C
(ii) < (i) < (iii) < (iv)

D
(iv) < (iii) < (i) = (ii)

Solution

10⁻⁴M NaCl i = 2
10⁻⁴M Urea i = 1
10⁻³M MgCl₂ i = 3
10⁻²M NaCl i = 2
More the value of i, C, more will be the elevation in boiling point hence increasing order of boding point is 10⁻⁴M Urea < 10⁻⁴M NaCl < 10⁻³M MgCl₂ < 10⁻²M NaCl
Question 6
4 / -1

Which of the following has the highest freezing point?

A
1 m NaCl solution

B
1 m KCl solution

C
1 m AlCl₃ solution

D
1 m C₆H₁₂O₆ solution

Solution

C₆H₁₂O₆ is a non-electrolyte hence furnishes minimum number of particles and will have maximum freezing point.
ΔT_f= iK_fm or ΔT_f ∝ i and ΔT_f= T^∘_f − T_f
Question 7
4 / -1

If α is the degree of dissociation of Na₂SO₄, the vant Hoff's factor (i) used for calculating the molecular mass is:

A
1 + α

B
1 − α

C
1 + 2α

D
1 − 2α

Solution
Question 8
4 / -1

For which of the following solutes the van’t Hoff factor is not greater than one?

A
NaNO₃

B
BaCl₂

C
K₄[Fe(CN)₆]

D
NH₂CONH₂

Solution

Urea is non-electrolyte, hence will not dissociate to give ions.
Question 9
4 / -1

What will be the degree of dissociation of 0.1M Mg(NO₃)₂ solution if van't Hoff factor is 2.74?

A
75%

B
87%

C
100%

D
92%

Solution

Mg(NO₃)₂ → Mg²⁺ + 2NO⁻₃

Degree of dissociation = 0.87 x 100 = 87%
Question 10
4 / -1

Which of the following will have the highest f.pt. at one atmosphere?

A
0.1 M NaCl solution

B
0.1 M sugar solution

C
0.1 M BaCl₂ solution

D
0.1 M FeCl₃ solution

Solution

For the same concentration of different solvents any colligative property ∝ = i
For NaCl, i = 2
Sugar solution, i = 1
BaCl₂, i = 3; FeCl₃, i = 4
Thus, for sugar solution depression in freezing point is minimum i.e., highest freezing point.
Question 11
4 / -1

A solute X when dissolved in a solvent associates to form a pentamer. The value of van't Hoff factor (i) for the solute will be

A
0.5

B
5

C
0.2

D
0.1

Solution

5A → A₅
Question 12
4 / -1

What will be the freezing point of a 0.5m KCl solution? The molal freezing point constant of water is 1.86^∘C m⁻¹.

A
−1.86^∘C

B
−0.372^∘C

C
−3.2^∘ C

D
0^∘C

Solution

ΔT_f= iK_f× m = 2 × 1.86 × 0.5 = 1.86^∘C
T_f = T^∘_f − ΔT_f= 0 − 1.86 = −1.86^∘C
Question 13
4 / -1

What amount of CaCl₂ (i = 2.47) is dissolved in 2 litres of water so that its osmotic pressure is 0.5atm at 27^oC?

A
3.42 g

B
9.24 g

C
2.834 g

D
1.820 g

Solution

Amount of CaCl₂ = n × M = 0.0164 × 111 = 1.820 g
Question 14
4 / -1

The van't Hoff factor of a 0.005 M aqueous solution of KCl is 1.95. The degree of ionisation of KCl is

A
0.95

B
0.97

C
0.94

D
0.93

Solution

KCl ⇔ K⁺+ Cl⁻ (n = 2)
Question 15
4 / -1

The elevation in boiling point of a solution of 9.43g of MgCl₂ in 1kg1kg of water is (Kb = 0.52 K kg mol⁻¹, Molar mass of MgCl₂ = 94.3 g mol⁻¹)

A
0.156

B
0.52

C
0.17

D
0.94

Solution

MgCl₂ ⇔ Mg²⁺ + 2Cl⁻, i = 3