Integer Answer Type Questions for JEE: Some Basic Concepts of Chemistry | Chapter-wise Tests for JEE Main & Advanced PDF Download

Q.1. Molar conductance of saturated BaSO₄ solution is 400ohm⁻¹cm² mol⁻¹ and its specific conductance is 4 × 10⁻⁵ ohm⁻¹ cm⁻¹. Hence solubility product value of BaSO₄  can be concluded as x ×10⁻⁸ M². Identify the x.

Ans. 1
Sp conduct = 4 x 10^-5
A = 1000 x 4 x 10^-5
= 4 x 10^-2
∴
K_sp = 10^-4 x 10^-4 = 10^-8 M²

Q.2. 10 ml of a gaseous hydrocarbon were burnt completely in 80 ml of O₂ at N.T.P. The remaining gas occupied 70 ml at N.T.P. This volume becomes 50 ml on treatment with KOH solution. What is the empirical formula weight of hydrocarbon?

Ans. 14
Let the molecular formula of hydrocarbon be C_xH_y.
C_xH_y +  (x + y/4)O₂ → xCO₂ +  y/2 H₂O.
Volume reacted 10 ml 10(x + y/4) - -
Volume produced 10x = 20 10x negligible
x = 2.
Since, the remaining gas occupied 70 ml,
∴ Total volume left = 70 = VO₂ + VCO₂
where VO₂ = volume of oxygen left and VCO₂ = Volume of CO₂ produced.
∴ VO₂  = 70 - 20 = 50 ml.
80 - 10(x + y/4) = 50
x + y/4 = 3
y/4 = 1
y = 4.
So, formula is C₂H₄.
Thus, empirical formula of the hydrocarbon is CH₂ and its empirical formula weight is 14.

Q.3. A solution is containing 2.52 g litre^-1 of a reductant. 25 ml of this solution required 20 ml of 0.01 M KMnO4 in acid medium for oxidation. Find the molecular weight of reductant. Given that each of the two atoms which undergo oxidation per molecule of reductant, show an increase in oxidation state by one unit.

Ans. 126
Meq. of reductant in 25 ml = Meq. of KMnO₄
= 20 × 0.01 × 5
∴ Meq. of reductant in 1 litre = 20 × 0.01 × 5 × 40 = 40.
Reductant shows the change A₂^+a → 2A^+b +  2e
∴ Equivalent weight of reductant = Molecular weight/2
∵ Meq. of reductant = 40


∴ M = 126.

Q.4. 1 gm of impure Na₂CO₃ is dissolved in water and solution made up to 250 ml. To 50 ml of this solution, 50 ml of 0.1 N HCl is added and mixture after shaking well, required 10 ml of 0.16 N NaOH solution for complete neutralization. Calculate % purity of sample of Na₂CO₃.

Ans. 90.1%
Let x gm be the weight of pure Na₂CO₃ in the sample.
Number of equivalents of Na₂CO₃ = x/53
Number of equivalents of Na₂CO₃ in 50 ml

Total meq. of HCl = 50 × 0.1 = 5
Meq. of HCl used by NaOH = 10 × 0.16 = 1.6
Meq. of HCl used by Na₂CO₃ = 5 - 1.6 = 3.4
∴ x/265 = 3.4 × 10^-3 ; x = 0.901 g
% purity of Na₂CO₃ = 0.901/1 × 100 = 90.1%.

Q.5. To a 25 mL H₂O₂ solution, excess of acidified solution of potassium iodide was added. The iodine liberated required 20 mL of 0.3 N sodium thiosulphate solution. Calculate the volume strength of H₂O₂ solution. Express your answer in millilitre.

Ans. 1344 ml.
2KI + H₂SO₄ + H₂O₂ → K₂SO₄ +2H₂O + I₂
2Na₂S₂O₃ + I₂ → Na₂S₄O₆ + 2NaI
Meq of Na₂S₂O₃ = 20 × 0.3 = 6
Meq of Na₂S₂O₃ ≡ Meq of I₂ = 6
Meq of I₂ ≡ Meq of H₂O₂ = 6
Wt. of H₂O₂ = Meq × Eq. wt. × 10^–3 = 6 × 17 × 10^–3 = 0.102 g
(∵ Eq. wt . of H₂O₂ = 34/2 = 17)
Strength of H₂O₂

Molarity of H₂O₂

2H₂O₂ → 2H₂O +  O₂
2 Moles       1 Mole
∴ 0.12 Moles 0.06 Mole
Volume of O₂ at STP = 0.06 × 22.4 Litre = 1.344 Litre
Hence, the volume strength of H₂O₂ = 1.344 Litre  =1344 ml.

Q.6. 0.56 gm sample of lime stone is dissolved and all Ca is precipitated as CaC₂O₄. These required 40 mL of 0.25 N KMnO₄ solution in acidic medium. Find the percentage of CaO in the lime stone.

Ans. 50%
Meq. of KMnO₄ =  40 × 0.25 = 10
Meq. of CaO = 10
gm eq. of CaO = 10 × 10^-3 = 1/100
Mass of CaO = 1/100 × 56/2 =0.28
% of CaO =  0.28/0.56 × 100 = 50%

Q.7. A definite amount of impure sample of P₄O₆ is treated with 20ml of X (M) KMnO₄ in acidic medium to produce H₃PO₄ and MnCl₂. 20 ml of same KMnO₄ on treatment with 0.2 M FeSO₄ requires exactly 10 ml of FeSO₄ solution. What is the amount of pure P₄O₆. If 0.1 g sample is taken calculate % purity of P₄O₆.

Ans. 55%
Mn⁺⁷ + 5e → Mn²⁺ (acidic medium)
P₄O₆ + KMnO₄ → H₃PO₄ + MnCl₂
Meq of KMnO₄ = Meq of P₄O₆
Meq of KMnO₄ = 10 × 0.2 × 1
or, 20 × 5 × X = 10 × 0.2 × 1
or, X = 0.02 M
∴ W/220 × 8 × 1000 = 20 × 0.02 × 5
W = 0.055 gm
% purity of P₄O₆ = 55%.

Q.8. 1 gm of impure Na₂CO₃ is dissolved in water and the solution is made upto 250 ml. To 50 ml of this made up solution, 50 ml of 0.1 N HCl is added and the mixture after shaking well, required 10 ml of 0.16 N NaOH solution for complete neutralization. Calculate % purity of the sample of Na₂CO₃. (Express your answer to the nearest whole number).

Ans. 90%.
Meq of HCl = 50 × 0.1 = 5
Meq of NaOH = 10 × 0.16 = 1.6
Meq of HCl = Meq of Na₂CO₃ + Meq of NaOH
Meq of Na₂CO₃ in 50 ml solution = 5 – 1.6 = 3.4
∴ In 250 ml Meq of Na₂CO₃ = (3.4 × 250) / 50 = 17
W/106 × 2 × 1000 = 17
W = 0.901 gm
% purity = 0.901/1 × 100 = 90.1 ≅ 90%.

Q.9. 1.5 g of a sample of ammonium sulphate was boiled with excess of NaOH solution. Evolved NH₃ was passed in 100 mL H₂SO₄. The partially neutralized acid required 160 mL N/2 NaOH for complete neutralization. Calculate % purity of ammonium sulphate ample.

Ans. 88

160 mL N/2 NaOH = 80 mL N NaOH ≡ 80 mL NH₂SO₄
∴ Consumed acid = 100 – 80 = 20 mL N H₂SO₄


2000 mL N H₂SO₄ ≡ 2 × 17 g NH₃ ≡ 132 g (NH₄)₂SO₄
∴ 20 mL NH₂SO₄ ≡ 132 × 20/2000 = 1.32 g (NH₄)₂SO₄
% (NH₄)₂SO₄ =  1.32/1.5 × 100 = 88.

Q.10. 20 ml solution of HIO₃ is reacted with excess of an aqueous solution of SO₂. The excess of SO₂ and I₂ formed are removed by heating the solution. The remaining solution in neutralized by 35.5 ml of 0.16 (N) NaOH solution. Calculate the strength of HIO₃ in gm/lt.
(Round off the answer to the nearest whole number)

Ans. 10 gm/lt.
20 × NHIO₃ = 35.5 × 0.16

N factor of HIO₃ = 5
Mass of HIO₃ = 0.284 × 176/5 = 10 gm/lt.