NEET Some Basic Concepts Of Chemistry — Set 10

Question 1

What is the mass of $ \ce{PbSO4} $ produced when 20.7 g of $ \ce{Pb(NO3)2} $ reacts with excess $ \ce{H2SO4} $? (Molar masses: $ \ce{Pb(NO3)2} $ = 331 g/mol, $ \ce{PbSO4} $ = 303 g/mol)

Accepted Answer

Reaction: $ \ce{Pb(NO3)2 + H2SO4 -> PbSO4 + 2HNO3} $. Moles of $ \ce{Pb(NO3)2} $ = $ \frac{20.7}{331} $ ≈ 0.0625 mol. 1 mol $ \ce{Pb(NO3)2} $ produces 1 mol $ \ce{PbSO4} $; mass = 0.0625 × 303 ≈ 18.94 g.

Question 2

What volume of $ \ce{CO2} $ at STP is produced when 15 g of $ \ce{C3H6} $ is burned completely? (Molar mass: $ \ce{C3H6} $ = 42 g/mol)

Accepted Answer

Reaction: $ \ce{2C3H6 + 9O2 -> 6CO2 + 6H2O} $. Moles of $ \ce{C3H6} $ = $ \frac{15}{42} $ ≈ 0.3571 mol. 2 mol $ \ce{C3H6} $ produce 6 mol $ \ce{CO2} $; 0.3571 mol produce $ \frac{6}{2} $ × 0.3571 ≈ 1.0714 mol. Volume = 1.0714 × 22.4 ≈ 24 L.

Question 3

A 0.78 g sample of a hydrocarbon produces 2.64 g of $ \ce{CO2} $ and 0.54 g of $ \ce{H2O} $ on complete combustion. What is its molecular formula if its molar mass is 78 g/mol? (Atomic masses: C = 12, H = 1, O = 16)

Accepted Answer

Mass of C = $ \frac{12}{44} $ × 2.64 ≈ 0.72 g; mass of H = $ \frac{2}{18} $ × 0.54 = 0.06 g. Total = 0.72 + 0.06 = 0.78 g (matches). Moles: C = $ \frac{0.72}{12} $ = 0.06, H = $ \frac{0.06}{1} $ = 0.06; ratio = 1 : 1; empirical formula = $ \ce{CH} $, mass = 13 g/mol. $ n = \frac{78}{13} = 6 $; molecular formula = $ \ce{C6H6} $.

NEET Chemistry: Some Basic Concepts Of Chemistry — Practice Set 10

Q1. What is the mass of \( \ce{PbSO4} \) produced when 20.7 g of \( \ce{Pb(NO3)2} \) reacts with excess \( \ce{H2SO4} \)? (Molar masses: \( \ce{Pb(NO3)2} \) = 331 g/mol, \( \ce{PbSO4} \) = 303 g/mol)

Q2. What volume of \( \ce{CO2} \) at STP is produced when 15 g of \( \ce{C3H6} \) is burned completely? (Molar mass: \( \ce{C3H6} \) = 42 g/mol)

Q3. A 0.78 g sample of a hydrocarbon produces 2.64 g of \( \ce{CO2} \) and 0.54 g of \( \ce{H2O} \) on complete combustion. What is its molecular formula if its molar mass is 78 g/mol? (Atomic masses: C = 12, H = 1, O = 16)

Q4. How many significant figures are present in the result of \( \frac{0.0567 \times 273.15}{0.0821} \)?

Q5. A solution of \( \ce{HCl} \) has a molarity of 2 M and a density of 1.06 g/mL. What is its molality? (Molar mass: \( \ce{HCl} \) = 36.5 g/mol)

Q6. How many grams of \( \ce{Al} \) are required to produce 5.6 L of \( \ce{H2} \) at STP with excess \( \ce{HCl} \)? (Atomic mass: Al = 27)

Q7. A solution contains 15 ppm of \( \ce{CaCl2} \) by mass in water. What is its molarity if the density is 1 g/mL? (Molar mass: \( \ce{CaCl2} \) = 111 g/mol)

Q8. What is the mass of \( \ce{PbCl2} \) produced when 27.8 g of \( \ce{Pb(NO3)2} \) reacts with excess \( \ce{NaCl} \)? (Molar masses: \( \ce{Pb(NO3)2} \) = 331 g/mol, \( \ce{PbCl2} \) = 278 g/mol)

Q9. What volume of \( \ce{CO2} \) at STP is produced when 12 g of \( \ce{C2H4} \) is burned completely? (Molar mass: \( \ce{C2H4} \) = 28 g/mol)

Q10. How many grams of \( \ce{Ca(OH)2} \) are required to produce 11.1 g of \( \ce{CaCl2} \) with excess \( \ce{HCl} \)? (Molar masses: \( \ce{Ca(OH)2} \) = 74 g/mol, \( \ce{CaCl2} \) = 111 g/mol)