JSMU Karachi Entry Test Chemistry Fundamental Concepts — Set 3

1. Question 1

   Q1. What is the molality of a solution with 12.0 g of urea (60.06 g/mol) in 250 g of water?

   • A) 0.800 m
   • B) 0.400 m
   • C) 2.00 m
   • D) 1.60 m

   Answer: 0.800 m

   Explanation: molality = (12/60.06)/0.250 = 0.800 m. B divides by 0.500 kg instead.

2. Question 2

   Q2. A 100.0 g sample of CuSO₄·5H₂O loses mass upon heating. What is the mass percent of H₂O? (CuSO₄ = 159.5 g/mol, H₂O = 18.02 g/mol)

   • A) 36.1%
   • B) 42.1%
   • C) 32.4%
   • D) 51.3%

   Answer: 36.1%

   Explanation: [(5×18.02)/(159.5+90.1)]×100 = 36.1%. B adds 18.02 five times incorrectly.

3. Question 3

   Q3. What is the mass of Mg in 6.0 g of Mg₃(PO₄)₂? (Mg = 24.31 g/mol, P = 30.97 g/mol, O = 16.00 g/mol)

   • A) 1.5 g
   • B) 2.2 g
   • C) 2.7 g
   • D) 3.0 g

   Answer: 2.7 g

   Explanation: Mass Mg = (3×24.31/262.86)×6.0 = 2.7 g. A calculates only 1 Mg atom.

4. Question 4

   Q4. A 25.0 mL sample of H₂SO₄ is titrated with 0.100 M NaOH. The endpoint is at 30.0 mL. What is the H₂SO₄ concentration?

   • A) 0.0600 M
   • B) 0.120 M
   • C) 0.240 M
   • D) 0.0300 M

   Answer: 0.0600 M

   Explanation: [2×0.100×0.030]/0.025 = 0.0600 M. B ignores 1:2 stoichiometry.

5. Question 5

   Q5. A gas mixture has 0.25 mol N₂ and 0.75 mol O₂. What is O₂’s partial pressure if total pressure is 1.0 atm?

   • A) 0.75 atm
   • B) 0.25 atm
   • C) 0.50 atm
   • D) 1.0 atm

   Answer: 0.75 atm

   Explanation: Partial pressure = 0.75/(0.25+0.75)×1.0 = 0.75 atm. C assumes equal moles.

6. Question 6

   Q6. What mass of CO₂ is produced from 10.0 g of C₂H₆ and excess O₂? (Molar mass C₂H₆ = 30.07 g/mol)

   • A) 29.3 g
   • B) 22.0 g
   • C) 14.7 g
   • D) 36.7 g

   Answer: 22.0 g

   Explanation: Molar ratio 2:4 → (10/30.07)×(44.01/30.07) = 22.0 g. C uses CO instead of CO₂.

7. Question 7

   Q7. A 50.0 mL solution is diluted to 200.0 mL. If final concentration is 0.150 M, what was the initial concentration?

   • A) 0.600 M
   • B) 0.300 M
   • C) 1.20 M
   • D) 0.450 M

   Answer: 0.600 M

   Explanation: C₁ = (0.150×200)/50 = 0.600 M. B assumes 1:1 volume ratio.

8. Question 8

   Q8. How many molecules are in 0.500 moles of NH₃? (Avogadro’s number = 6.022×10²³)

   • A) 3.01×10²³
   • B) 6.02×10²³
   • C) 1.20×10²⁴
   • D) 9.03×10²³

   Answer: 3.01×10²³

   Explanation: 0.500×6.022×10²³ = 3.01×10²³. C uses 2×Avogadro’s number.

9. Question 9

   Q9. What is the normality of 0.250 M H₃PO₄ for reaction with NaOH to form NaH₂PO₄?

   • A) 0.250 N
   • B) 0.500 N
   • C) 0.750 N
   • D) 1.00 N

   Answer: 0.500 N

   Explanation: Normality = 0.250×2 = 0.500 N (1 H+ transferred). C assumes 3 H+.

10. Question 10

    Q10. A 15.0 g sample of NaNO₃ is dissolved in 500. g of water. What is the solution’s concentration in ppm?

    • A) 30 ppm
    • B) 300 ppm
    • C) 6000 ppm
    • D) 150 ppm

    Answer: 30 ppm

    Explanation: ppm = (15.0/515.0)×10⁶ ≈ 30 ppm. B assumes 500 g instead of 515 g total.

11. Question 11

    Q11. What is the molarity of 12.0 M HCl diluted to 500.0 mL from 100.0 mL?

    • A) 2.40 M
    • B) 1.20 M
    • C) 6.00 M
    • D) 3.00 M

    Answer: 2.40 M

    Explanation: M₁V₁ = 12×0.1 = 1.2 → M₂ = 1.2/0.5 = 2.40 M. B uses 0.05 L instead of 0.5 L.

12. Question 12

    Q12. A 20.0 g sample of KBr is in 100. g of solution. What is the mass percent of KBr?

    • A) 16.7%
    • B) 20.0%
    • C) 18.6%
    • D) 25.0%

    Answer: 20.0%

    Explanation: Mass% = (20.0/100.0)×100 = 20.0%. A assumes 120.0 g total mass.

13. Question 13

    Q13. What volume of 0.500 M AgNO₃ is needed to precipitate all Cl⁻ in 25.0 mL of 0.100 M MgCl₂?

    • A) 10.0 mL
    • B) 20.0 mL
    • C) 5.00 mL
    • D) 25.0 mL

    Answer: 10.0 mL

    Explanation: Moles Cl⁻ = 0.025×0.100×2 = 0.005 → Volume = 0.005/(0.500) = 0.010 L. B uses 1:1 ratio.

14. Question 14

    Q14. What is the molarity of a solution containing 4.0 g of NaOH (molar mass = 40 g/mol) dissolved in 500 mL of solution?

    • A) 0.1 M
    • B) 0.2 M
    • C) 0.3 M
    • D) 0.4 M

    Answer: 0.2 M

    Explanation: Molarity = moles/volume (L). 4.0g/40g/mol = 0.1 mol. 500 mL = 0.5 L. 0.1/0.5 = 0.2 M. Option A fails due to dividing 4.0g by 500.

15. Question 15

    Q15. A 250 mL solution of 0.5 M HCl is diluted to 500 mL. What is the new concentration?

    • A) 0.125 M
    • B) 0.25 M
    • C) 0.50 M
    • D) 1.00 M

    Answer: 0.25 M

    Explanation: C1V1 = C2V2. 0.5 × 250 = C2 × 500 → C2 = 0.25 M. Option D assumes doubling volume halves concentration incorrectly.

16. Question 16

    Q16. How many moles of O2 are produced when 4 moles of H2O2 decompose (2H2O2 → 2H2O + O2)?

    • A) 1 mole
    • B) 2 moles
    • C) 3 moles
    • D) 4 moles

    Answer: 2 moles

    Explanation: 2 moles H2O2 produce 1 mole O2. 4 moles H2O2 produce 2 moles O2. Option D assumes 1:1 ratio.

17. Question 17

    Q17. What is the mass percent of carbon in CO2? (C = 12, O = 16)

    • A) 27.27%
    • B) 33.00%
    • C) 44.00%
    • D) 25.00%

    Answer: 27.27%

    Explanation: Mass percent = (12/44) × 100 = 27.27%. Option B adds O2 mass instead of CO2.

18. Question 18

    Q18. 24 g of Mg (atomic mass = 24) combines with 16 g of O (atomic mass = 16). What is the empirical formula?

    • A) MgO2
    • B) MgO
    • C) Mg2O
    • D) Mg2O3

    Answer: MgO

    Explanation: Mole ratio Mg:O = 1:1. Option A assumes O:O ratio incorrectly.

19. Question 19

    Q19. If 0.5 mol of O2 occupies 11.2 L at STP, what pressure is exerted? (R = 0.0821 L·atm/mol·K)

    • A) 1.0 atm
    • B) 2.0 atm
    • C) 0.5 atm
    • D) 4.0 atm

    Answer: 1.0 atm

    Explanation: PV = nRT → P = (0.5 × 0.0821 × 273)/11.2 ≈ 1 atm. Option B assumes 22.4 L instead of 11.2 L.

20. Question 20

    Q20. In the reaction 2Al + 3Cl2 → 2AlCl3, which reactant is limiting if 1 mole of Al and 1.5 moles of Cl2 are used?

    • A) Al
    • B) Cl2
    • C) Neither
    • D) Both

    Answer: Neither

    Explanation: Mole ratio 1:1.5 matches 2:3. Both reactants fully consume. Option A assumes a different ratio.