10th Science Expected Important Problems 2026

Conquer the trickiest parts of your exam with our master list of 10th Science Expected Important Problems 2026. Numerical problems in Physics and Chemistry—such as lens formulas, electrical resistance networks, and chemical mole concepts—often determine the difference between a good score and a perfect 100%. This page features a highly curated selection of step-by-step solved problems aligned with the latest 2026 exam blueprint, ensuring you know exactly which formulas to apply and how to present your steps to secure full marks.

Expected Chemistry Problems – 2026 (Q.No: 32)

1. Find the percentage of nitrogen in ammonia. (N-14, H-1) (PTA – 1)

Molar mass of NH₃ = 14 + 3 = 17 g
Mass % of Nitrogen = (14 / 17) × 100 = 82.35%

2. Calculate the number of moles in (PTA - 5)
i) 27g of Al (JUL - 24) ii) 1.51 × 10²³ molecules of NH₄Cl (APR - 24)

i) 27g of Al:
Number of moles = Mass / Atomic mass = 27 / 27 = 1 mole

ii) 1.51 × 10²³ molecules of NH₄Cl:
Number of moles = Number of Molecules / Avogadro's number = (1.51 × 10²³) / (6.023 × 10²³) = 1 / 4 = 0.25 mole

3. Calculate the gram molecular mass of the following: 1) H₂O 2) CO₂ 3) Ca₃(PO₄)₂

1) H₂O (Atomic masses of H = 1, O = 16):
Gram molecular mass of H₂O = (1 × 2) + (16 × 1) = 2 + 16 = 18 g

2) CO₂: (Atomic masses of C = 12, O = 16):
Gram molecular mass of CO₂ = (12 × 1) + (16 × 2) = 12 + 32 = 44 g

3) Ca₃(PO₄)₂: (Atomic masses of Ca = 40, P = 30, O = 16):
Gram molecular mass of Ca₃(PO₄)₂ = (40 × 3) + [30 + (16 × 4)] × 2 = 120 + (94 × 2) = 120 + 188 = 308 g

4. Calculate the number of moles in 46 g of sodium?

Number of moles = Mass / Atomic mass = 46 / 23 = 2 mole

5. Calculate the number of moles of a sample that contains 12.046 × 10²³ atoms of iron?

Number of moles = Number of Atoms / Avogadro's number = (12.046 × 10²³) / (6.023 × 10²³) = 2 mole

6. Calculate % of S in H₂SO₄

Molecular mass of H₂SO₄ = (1 × 2) + (32 × 1) + (16 × 4) = 2 + 32 + 64 = 98 g
% of S in H₂SO₄ = (Mass of sulphur / Molecular mass) × 100 = (32 / 98) × 100 = 32.65%

7. A is a reddish brown metal, which combines with O₂ at < 1370 K gives B, a black coloured compound. At a temperature > 1370 K, A gives C which is red in colour. Find A, B and C with reaction. (PTA - 4)

Copper is a reddish brown metal (A).
Cu (A) combines with O₂ at < 1370 K gives a black coloured cupric oxide (B):
2Cu + O₂ →_{(< 1370 K)} 2CuO (B) [Cupric oxide (black colour)]
Cu (A) reacts with O₂ at > 1370 K to give cuprous oxide (C):
4Cu + O₂ →_{(> 1370 K)} 2Cu₂O (C) [(Cuprous oxide) red colour]

A	Cu	Copper
B	CuO	Cupric oxide
C	Cu₂O	Cuprous oxide

8. A is a silvery white metal. A combines with O₂ to form B at 800°C the alloy of A is used in making the aircraft. Find A and B (PTA-1)

Al is a silvery white metal (A)
'Al' combines with O₂ to form Aluminium oxide - Al₂O₃ (B) at 800°C:
4Al (A) + 3O₂ →_(800°C) 2Al₂O₃ (B) (Aluminium oxide)
Aluminium Alloys are used in making the aircraft, tools, pressure cookers, instruments etc.

A	Al	Aluminium
B	Al₂O₃	Aluminium oxide

9. A solution is prepared by dissolving 45 g of sugar in 180 g of water. Calculate the mass percentage of solute.

Mass of solute = 45 g
Mass of solvent = 180 g
Mass percentage of solute = [Mass of solute / (Mass of solvent + Mass of solute)] × 100
= [45 / (180 + 45)] × 100 = (4500 / 225) = 20%
Mass percentage of solute = 20%

10. 3.5 litres of ethanol is present in 15 litres of aqueous solution of ethanol. Calculate volume percent of ethanol solution.

Volume of solute = 3.5 litres
Volume of solution = 15 litres
Volume percentage = (Volume of solute / Volume of solution) × 100 = (3.5 / 15) × 100 = 23.33%
Volume percentage of ethanol solution is = 23.33%

11. A solution was prepared by dissolving 25 g of sugar in 100 g of water. Calculate the mass percentage of solute.

Mass of the solute = 25 g
Mass of the solvent = 100 g
Mass percentage = [Mass of the solute / (Mass of the solute + Mass of the solvent)] × 100
= [25 / (25 + 100)] × 100 = (25 / 125) × 100 = 20%
Mass percentage = 20%

12. A solution is made from 35 ml of Methanol and 65 ml of water. Calculate the volume percentage.

Volume of the solute (methanol) = 35 ml
Volume of the solvent (water) = 65 ml
Volume percentage = [Volume of the solute / Volume of the solution] × 100
Volume percentage = [Volume of the solute / (Volume of the solute + Volume of the solvent)] × 100
Volume percentage = [35 / (35 + 65)] × 100 = (35 / 100) × 100 = 35%

13. A solution was prepared by dissolving 25 grams of sugar in 100g of water. Calculate the mass percentage of the solute.

Mass of the solute = 25 g
Mass of the solvent = 100 g
Mass percentage of solute = [Mass of the solute / (Mass of the solute + Mass of the solvent)] × 100
= [25 / (25 + 100)] × 100 = (2500 / 125) = 20%
Mass percentage of solute = 20%

14. Lemon juice has a pH 2, what is the concentration of H⁺ ions?

pH = 2
[H⁺] = 1.0 × 10^-2
The concentration of H⁺ ion is 1.0 × 10^-2 mole litre^-1

15. Calculate the pH of 1.0 × 10^-4 molar solution of HNO₃.

[H⁺] = 1.0 × 10^-4
pH = -log₁₀[H⁺]
= -log₁₀[1.0 × 10^-4]
= -[log₁₀1 + log₁₀10^-4]
= -[log₁₀1 - 4 log₁₀10]
pH = -[0 - (4 × 1)] = -[-4]
pH = +4

16. What is the pH of 1.0 × 10^-5 molar solution of KOH?

[OH^-] = 1.0 × 10^-5
pOH = -log₁₀[OH^-]
= -log₁₀[1 × 10^-5]
= -[log₁₀1 - 5 log₁₀10]
pOH = -[0 - (5 × 1)]
pOH = 5
Since pH + pOH = 14
pH = 14 - pOH = 14 - 5
pH = 9

17. The hydroxide ion concentration of a solution is 1 × 10^-11 M. What is the pH of the solution?

[OH^-] = 1.0 × 10^-11
pOH = -log₁₀[OH^-]
= -log₁₀[1 × 10^-11]
= -[log₁₀1 - 11 log₁₀10]
pOH = -[0 - (11 × 1)]
pOH = +11
Since pH + pOH = 14
pH = 14 - pOH = 14 - 11
pH = 3

18. Calculate the pH of 1.0 × 10^-5 solution of KOH.

[OH^-] = 1.0 × 10^-5 mol litre^-1
pOH = -log₁₀[OH^-]
= -log₁₀[1 × 10^-5]
= -[log₁₀1 - 5 log₁₀10]
pOH = -[0 - (5 × 1)]
pOH = 5
Since pH + pOH = 14
pH = 14 - pOH = 14 - 5
pH = 9

19. Calculate the pH of 1 × 10^-4 molar solution of NaOH.

[OH^-] = 1.0 × 10^-4
pOH = -log₁₀[OH^-]
= -log₁₀[1 × 10^-4]
= -[log₁₀1 - 4 log₁₀10]
pOH = -[0 - (4 × 1)]
pOH = 4
Since pH + pOH = 14
pH = 14 - pOH = 14 - 4
pH = 10