The molarity of the solution containing 7.1 g of Na₂SO₄ in 100 ml of aqueous solution is

2 M
0.5 M
1 M
0.05 M

Solution

An X molal solution of a compound in benzene has mole fraction of solute equal to 0.2. The value of X is

Concentrated hydrochloric acid contains 37% HCl (by mass).The density of this solution is 1.18 gm/ml. The molarity of the solution is :

10 M
12 M
13 M
14 M

Solution

The molarity of a solution containing 5 gm of NaOH in 250 ml solution :

Solution

Molarity of liquid HCl will be, if density of solution is 1.17 gm/cc

36.5
32.05
18.25
42.10

Solution

Which of the following modes of expressing concentration is independent of temperature ?

Molarity
Molality
Formality
Normality

Solution

The molality involves weights of the solute and the solvent. Since weight does not change with the temperature, therefore molality does not depend upon the temperature.

The statement “If 0.003 moles of a gas are dissolved in 900 g of water under a pressure of 1 atmosphere, 0.006 moles will be dissolved under a pressure of 2 atmospheres”, illustrates

Dalton’s law of partial pressure
Graham’s law
Raoult’s law
Henry’s law

An aqueous solution containing 1 gm of urea boils at 100.25ºC. The aqueous solution containing 3 g of glucose in the same volume will boil at (mol. wt. of urea = 60 and of glucose= 180) :

100.25ºC
100ºC
100.15ºC
100.50ºC

Solution

Formation of a solution from two components can be considered as
(i)Pure solvent ⟶separated solvent molecules, ΔH₁
(ii)Pure solute ⟶separated solute molecules, Δ H₂
(iii)Separated solvent &solute molecules ⟶ Solution,ΔH₃ Solution so formed will be ideal if

ΔH_soln = ΔH₃ - ΔH₁ - ΔH₂
ΔH_soln = ΔH₁ + ΔH₂ + ΔH₃
ΔH_soln = ΔH₁ + ΔH₂ - ΔH₃
ΔH_soln = ΔH₁ - ΔH₂ - ΔH₃

Solution

For an ideal solution, ΔH_mixing = 0
ΔH = ΔH₁ + ΔH₂ + ΔH₃ (According to Hess's law)i.e., for ideal solutions there is no change in magnitude of the attractive forces in the two components present.

Lowering in vapur pressrue is the highest for :

0.1 M glucose
0.1 M BaCl₂
0.2 M urea
0.1 M MgSO₄

Solution

Lowering in vapour pressure is the highest for that molecules for which Vont-Hoff factor is maximum be-cause. Now, Von’t Hoff factor is maximum for BaCl2 [As number of ions formed here after dissociation in water is zero], this will have the highest lowering of vapour pressure.

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