BSTI Previous Exam Question Pattern

Life's Canvas with Reality
11 Jan, 2025

 BSTI Job Exam Questions Pattern

  1. What is the molarity of a solution prepared by dissolving 10.2 g of glucose (C₆H₁₂O₆) in 414 mL of solution?

  2. Calculate the molality of a solution made by adding 9.5 g of NaCl to 300 g of water.

  3. How many milliliters of a 5.0 M CuSO₄ solution are needed to prepare 0.350 L of 0.500 M CuSO₄?

  4. A solution is 15.2% by mass hydrogen peroxide (H₂O₂) with a density of 1.01 g/mL. What is its molarity?

  5. What is the normality of a 0.5 M H₂SO₄ solution?

  6. Determine the molarity of a solution containing 211 g of sodium hydrogen carbonate (NaHCO₃) dissolved in 10.0 L of solution.

  7. What is the molality of a solution prepared by dissolving 58.5 g of NaCl in 500 g of water?

  8. How many grams of KOH are needed to prepare 2.0 L of a 0.5 M solution?

  9. Calculate the molarity of a solution containing 30.5 g of glucose (C₆H₁₂O₆) in 750 mL of solution.

  10. What is the molality of a solution made by dissolving 20 g of ethanol (C₂H₅OH) in 250 g of water?

  11. How many milliliters of a 12 M HCl solution are required to prepare 1.0 L of a 1.0 M HCl solution?

  12. A solution contains 5.0 g of acetic acid (CH₃COOH) in 100 g of water. Calculate its molality.

  13. What is the normality of a 1.0 M solution of phosphoric acid (H₃PO₄)?

  14. Determine the molarity of a solution prepared by dissolving 85 g of KNO₃ in enough water to make 1.5 L of solution.

  15. Calculate the molality of a solution containing 10 g of urea (CH₄N₂O) dissolved in 200 g of water.

  16. How many grams of NaOH are required to prepare 500 mL of a 2 M solution?

  17. What is the molarity of a solution containing 40 g of NaCl in 1.0 L of solution?

  18. Calculate the molality of a solution made by dissolving 75 g of glucose (C₆H₁₂O₆) in 400 g of water.

  19. How many milliliters of a 6 M H₂SO₄ solution are needed to prepare 2.0 L of a 0.5 M solution?

  20. A solution is prepared by dissolving 10 g of NaOH in 250 mL of solution. What is its molarity?



The Answers of the Questions

Q1. 

Molarity Calculation: 
Moles of glucose: 
Molar mass of C₆H₁₂O₆ = 180.16 g/mol.
Moles = 10.2 g / 180.16 g/mol = 0.0566 mol.
Volume in liters: 414 mL = 0.414 L.
Molarity (M): M = moles/volume = 0.0566 mol / 0.414 L ≈ 0.137 M.

Q2

Molality Calculation: 
Moles of NaCl: Molar mass of NaCl = 58.44 g/mol.
Moles = 9.5 g / 58.44 g/mol = 0.1625 mol.
Mass of solvent in kg: 300 g = 0.300 kg.
Molality (m): m = moles / mass = 0.1625 mol / 0.300 kg ≈ 0.542 m.

Q3

Dilution Calculation:


Using M₁V₁ = M₂V₂:
(5.0 M) × V₁ = (0.5 M) × (0.350 L)
V₁ = (0.5 M × 0.350 L) / 5.0 M = 0.035 L = 35 mL.


Q4

Molarity Calculation:

Assume 1 L of solution:


Mass = 1.01 g/mL × 1000 mL = 1010 g.
Mass of H₂O₂:
15.2% of 1010 g = 0.152 × 1010 g = 153.52 g.
Moles of H₂O₂: Molar mass = 34.02 g/mol.
Moles = 153.52 g / 34.02 g/mol ≈ 4.51 mol.
Molarity (M): M = moles / volume = 4.51 mol / 1 L = 4.51 M.


  1. Normality of 0.5 M H₂SO₄:

    • H₂SO₄ provides 2 moles of H⁺ ions per mole of H₂SO₄.
    • Normality (N) = Molarity × Number of equivalents = 0.5 M × 2 = 1.0 N.

  2. Molarity of NaHCO₃ solution:

    • Molar mass of NaHCO₃ = 84.01 g/mol.
    • Moles of NaHCO₃ = 211 g / 84.01 g/mol = 2.51 mol.
    • Volume of solution = 10.0 L.
    • Molarity (M) = Moles / Volume = 2.51 mol / 10.0 L = 0.251 M.

  3. Molality of NaCl solution:

    • Molar mass of NaCl = 58.44 g/mol.
    • Moles of NaCl = 58.5 g / 58.44 g/mol = 1.0 mol.
    • Mass of water = 500 g = 0.500 kg.
    • Molality (m) = Moles / Mass = 1.0 mol / 0.500 kg = 2.0 m.

  4. Grams of KOH for 2.0 L of 0.5 M solution:

    • Molar mass of KOH = 56.11 g/mol.
    • Moles of KOH = Molarity × Volume = 0.5 mol/L × 2.0 L = 1.0 mol.
    • Mass = Moles × Molar mass = 1.0 mol × 56.11 g/mol = 56.11 g.

  5. Molarity of glucose solution:

    • Molar mass of glucose = 180.16 g/mol.
    • Moles of glucose = 30.5 g / 180.16 g/mol = 0.1693 mol.
    • Volume in liters = 750 mL = 0.750 L.
    • Molarity (M) = Moles / Volume = 0.1693 mol / 0.750 L = 0.226 M.

  6. Molality of ethanol solution:

    • Molar mass of ethanol = 46.07 g/mol.
    • Moles of ethanol = 20 g / 46.07 g/mol = 0.434 mol.
    • Mass of water = 250 g = 0.250 kg.
    • Molality (m) = Moles / Mass = 0.434 mol / 0.250 kg = 1.736 m.

  7. Volume of 12 M HCl to prepare 1.0 L of 1.0 M solution:

    • Using M1V1=M2V2M_1V_1 = M_2V_2:
    • (12M)⋅V1=(1.0M)⋅(1.0L)(12 M) \cdot V_1 = (1.0 M) \cdot (1.0 L)
    • V1=1.0L/12=0.0833L=83.3 mL.V_1 = 1.0 L / 12 = 0.0833 L = 83.3 \, \text{mL}.

  8. Molality of acetic acid solution:

    • Molar mass of acetic acid = 60.05 g/mol.
    • Moles of acetic acid = 5.0 g / 60.05 g/mol = 0.0833 mol.
    • Mass of water = 100 g = 0.100 kg.
    • Molality (m) = Moles / Mass = 0.0833 mol / 0.100 kg = 0.833 m.

  9. Normality of 1.0 M H₃PO₄:

    • H₃PO₄ dissociates to provide 3 H⁺ ions.
    • Normality (N) = Molarity × Number of equivalents = 1.0 M × 3 = 3.0 N.

  10. Molarity of KNO₃ solution:

    • Molar mass of KNO₃ = 101.11 g/mol.
    • Moles of KNO₃ = 85 g / 101.11 g/mol = 0.841 mol.
    • Volume in liters = 1.5 L.
    • Molarity (M) = Moles / Volume = 0.841 mol / 1.5 L = 0.561 M.

  11. Molality of urea solution:

    • Molar mass of urea = 60.06 g/mol.
    • Moles of urea = 10 g / 60.06 g/mol = 0.166 mol.
    • Mass of water = 200 g = 0.200 kg.
    • Molality (m) = Moles / Mass = 0.166 mol / 0.200 kg = 0.830 m.

  12. Grams of NaOH for 500 mL of 2 M solution:

    • Molar mass of NaOH = 40.00 g/mol.
    • Moles of NaOH = Molarity × Volume = 2 mol/L × 0.5 L = 1.0 mol.
    • Mass = Moles × Molar mass = 1.0 mol × 40.00 g/mol = 40.0 g.

  13. Molarity of NaCl solution:

    • Molar mass of NaCl = 58.44 g/mol.
    • Moles of NaCl = 40 g / 58.44 g/mol = 0.684 mol.
    • Volume = 1.0 L.
    • Molarity (M) = Moles / Volume = 0.684 mol / 1.0 L = 0.684 M.

  14. Molality of glucose solution:

    • Molar mass of glucose = 180.16 g/mol.
    • Moles of glucose = 75 g / 180.16 g/mol = 0.416 mol.
    • Mass of water = 400 g = 0.400 kg.
    • Molality (m) = Moles / Mass = 0.416 mol / 0.400 kg = 1.04 m.

  15. Volume of 6 M H₂SO₄ for 2.0 L of 0.5 M solution:

    • Using M1V1=M2V2M_1V_1 = M_2V_2:
    • (6M)⋅V1=(0.5M)⋅(2.0L)(6 M) \cdot V_1 = (0.5 M) \cdot (2.0 L)
    • V1=(0.5M×2.0L)/6M=0.167L=167 mL.V_1 = (0.5 M × 2.0 L) / 6 M = 0.167 L = 167 \, \text{mL}.

  16. Molarity of NaOH solution:

    • Molar mass of NaOH = 40.00 g/mol.
    • Moles of NaOH = 10 g / 40.00 g/mol = 0.25 mol.
    • Volume in liters = 250 mL = 0.250 L.
    • Molarity (M) = Moles / Volume = 0.25 mol / 0.250 L = 1.0 M.

This is the written question pattern for the chemical wing at BSTI.

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