JEE Advanced (Subjective Type Questions): Solutions- 1 | Chapter-wise Tests for JEE Main & Advanced PDF Download

 1. What is the molarity and molality of a 13% solution (by weight) of sulphuric acid with a density of 1.02 g/ml? To what volume should 100 ml of this acid be diluted in order to prepare a 1.5 N solution?          (1978)

Ans : 1.38 M, 1.57 m, 184 ml

Solution : 
 TIPS/Formulae :  (i) 

 (ii) N₁V₁ = N₂V₂ 

A 13% solution (by weight) contains 13g of solute (i.e. H₂SO₄) per 100 gm of solution

Volume of solution in L

Mass of solute in 100 ml of solution = 13 g [ 13 % solution]
Mass of solvent = Mass of solution – Mass of solvent
= 100 – 13 = 87 g

N₁ = 2.70, V₁ = 100 ml, N₂ = 1.5, V₂ = ?             

N₁V₁ = N₂V₂; 2.70 × 100 = 1.5 × V₂

∴ 100 ml of this acid should be diluted to 180 ml to prepare 1.5 N solution

2. A bottle of commercial sulphuric acid (density 1.787 g/ml) is labelled as 86 percent by weight. What is the molarity of the acid. What volume of the acid has to be used to make 1 litre of 0.2 M H₂SO₄?       (1979)

Ans : 15.81M; 12.65

Solution : 
 

M₁V₁ = M₂V₂
M₁ = 15.81, V₁ = ?
M₂ = 0.2, V₂ = 1 L = 1000 ml
∴ 15.81 × V₁ = 0.2 × 1000

∴ Amount of acid to be used to make 1 L of 0.2 M H₂SO₄ = 12.65.
 

3. 0.5 gm of fuming H₂SO₄ (Oleum) is diluted with water. This solution is completely neutralized by 26.7 ml of 0.4 N NaOH. Find the percentage of free SO₃ in the sample of oleum.         (1980)

Ans : 3.84%

Solution : 

49g (Q eq wt of H₂SO₄ = 49) of H₂SO₄ will be neutralised by = 1N 1000 ml NaOH
∴ 0.5g of H2SO4 will be neutralised by

Volume of 1 N NaOH used by dissolved
SO₃ = 10.68 – 10.20 = 0.48 ml
SO₃ + 2NaOH →   Na₂SO₄ + H₂O

Wt of SO₃ in 0.48 ml of 1 M solution

4. The vapour pressure of pure benzene is 639.7 mm of mercury and the vapour of a solution of a solute in benzene at the same temperature is 631.9 mm of mercury. Calculate the molality of the solution. (1981 - 3 Marks)

Ans : 0.156 m/kg

Solution : 

TIPS/Formulae : 

Now the solution contains ‘m’ moles of solute per 1000 gm of benzene
Vapour pressure of benzene, P^o = 639.7 mm
Vapour pressure of solution, P = 631.9 mm
Moles of benzene (Mol. wt. 78), N = 1000/ 78
Moles of solute, n = ?
Substitute these values in the Raoult’s equation

 Hence, molality of solution = 0.156 m

5. An organic compound CxH2yOy was burnt with twice the amount of oxygen needed for complete combustion to CO₂ and H2O. The hot gases when cooled to 0ºC and 1 atm. pressure, measured 2.24 liters. The water collected during cooling weighed 0.9 g. The vapour pressure of pure water at 20ºC is 17.5 mm Hg and is lowered by 0.104 mm when 50 g of the organic compound are dissolved in 1000 g of water. Give the molecular formula of the organic compound. (1983 - 5 Marks)

Ans : (CH₂O)₅ or C₅H₁₀O₅

Solution :  The chemical equation for the combustion of organic compound C_xH_2yO_y can be represented as :
C_xH_2yO_y + 2xO₂ = x CO₂ + y H₂O + x O₂
The gases obtained after cooling = x + x = 2x
∴ 2x = 2.24 litres                               [∵ H₂O is in liquid state]

Number of moles of CO₂ =  

=1/20 mole = 0.05 mole

The empirical formula of the organic compound is C(H₂O) ...(1)
The mole fraction of the solute (A)
= relative decrease in vapour pressure of the solvent (B)

Molecular wt. of the organic compound
(CH₂O)n = 150
Molecular wt. of CH₂O = 12 + 2 + 16 = 30

∴ 30 × n = 150                  [∵(CH₂O)n = mol. formula]
or n = 150/30 = 5

∴ Molecular formula of the given organic compound is
(CH₂O)₅ or C₅H₁₀O₅.
 

6. 'Two volatile and miscible liquids can be separated by fractional distillation into pure component', is true under what conditions? (1984 - 1 Mark)

Solution : If they form an ideal solution which obeys’ Raoult’s Law and for which
ΔH_mixing = 0 and ΔV_mixing = 0
Thus we can separate two volatile and miscible liquids by fractional distillation if, they should not form azeotropic solutions.