NEET Some Basic Concepts Of Chemistry — Set 3

Question 1

A 0.25 M $ \ce{NaOH} $ solution is diluted by adding 300 mL of water to 200 mL of the original solution. What is the final molarity?

Accepted Answer

$ M_1 V_1 = M_2 V_2 $. 0.25 × 200 = $ M_2 $ × (200 + 300). $ M_2 = \frac{0.25 \times 200}{500} = 0.1 $ M.

Question 2

A gas mixture contains 2 g of $ \ce{H2} $ and 16 g of $ \ce{O2} $. What is the total number of molecules? (Molar masses: $ \ce{H2} $ = 2 g/mol, $ \ce{O2} $ = 32 g/mol, Avogadro number = $ 6.022 \times 10^{23} $)

Accepted Answer

Moles of $ \ce{H2} $ = $ \frac{2}{2} $ = 1 mol; moles of $ \ce{O2} $ = $ \frac{16}{32} $ = 0.5 mol. Total moles = 1 + 0.5 = 1.5 mol. Molecules = 1.5 × $ 6.022 \times 10^{23} $ = $ 9.033 \times 10^{23} $.

Question 3

A compound contains 52.17% carbon, 13.04% hydrogen, and 34.78% oxygen by mass. What is its empirical formula? (Atomic masses: C = 12, H = 1, O = 16)

Accepted Answer

For 100 g: C = 52.17 g, H = 13.04 g, O = 34.78 g. Moles: C = $ \frac{52.17}{12} $ ≈ 4.35, H = $ \frac{13.04}{1} $ ≈ 13.04, O = $ \frac{34.78}{16} $ ≈ 2.17. Ratio = 2 : 6 : 1; empirical formula = $ \ce{C2H6O} $.

NEET Chemistry: Some Basic Concepts Of Chemistry — Practice Set 3

Q1. A 0.25 M \( \ce{NaOH} \) solution is diluted by adding 300 mL of water to 200 mL of the original solution. What is the final molarity?

Q2. A gas mixture contains 2 g of \( \ce{H2} \) and 16 g of \( \ce{O2} \). What is the total number of molecules? (Molar masses: \( \ce{H2} \) = 2 g/mol, \( \ce{O2} \) = 32 g/mol, Avogadro number = \( 6.022 \times 10^{23} \))

Q3. A compound contains 52.17% carbon, 13.04% hydrogen, and 34.78% oxygen by mass. What is its empirical formula? (Atomic masses: C = 12, H = 1, O = 16)

Q4. How many grams of \( \ce{Al2O3} \) are required to produce 5.4 g of \( \ce{Al} \) with excess \( \ce{CO} \)? (Molar masses: \( \ce{Al2O3} \) = 102 g/mol, Al = 27 g/mol)

Q5. What volume of \( \ce{O2} \) at STP is required to burn 6 g of \( \ce{CH4} \) completely to \( \ce{CO2} \) and \( \ce{H2O} \)? (Molar mass: \( \ce{CH4} \) = 16 g/mol)

Q6. What is the mass percentage of \( \ce{NH3} \) in a solution made by dissolving 3.4 g of \( \ce{NH3} \) in 16.6 g of water? (Molar mass: \( \ce{NH3} \) = 17 g/mol)

Q7. A 0.56 g sample of a hydrocarbon produces 1.76 g of \( \ce{CO2} \) and 0.72 g of \( \ce{H2O} \) on complete combustion. What is its molecular formula if its molar mass is 56 g/mol? (Atomic masses: C = 12, H = 1, O = 16)

Q8. How many significant figures are present in the result of \( \frac{0.0678 \times 310.15}{0.0821} \)?

Q9. What is the mass of \( \ce{AlCl3} \) produced when 13.5 g of \( \ce{Al} \) reacts with excess \( \ce{Cl2} \)? (Molar masses: Al = 27 g/mol, \( \ce{AlCl3} \) = 133.5 g/mol)

Q10. What is the mass of \( \ce{MgSO4} \) produced when 8.4 g of \( \ce{Mg(OH)2} \) reacts with excess \( \ce{H2SO4} \)? (Molar masses: \( \ce{Mg(OH)2} \) = 58 g/mol, \( \ce{MgSO4} \) = 120 g/mol)