Mole Concept — Set 4
Chemistry · मोल संकल्पना · Questions 31–40 of 40
What is the molarity of a solution containing 5.85 g NaCl in 1.0 L of solution? (NaCl = 58.5 g/mol)?
Correct Answer: B. 0.10 M
Option B is correct because moles of NaCl = 5.85/58.5 = 0.10 mol. Molarity = moles/volume = 0.10/1.0 = 0.10 M. This uses molar mass to convert grams into moles.
How much 10.0 M stock solution is needed to prepare 250 mL of 1.0 M solution?
Correct Answer: B. 25 mL
Option B is correct because M1V1 = M2V2 gives V1 = (1.0 × 0.250)/10.0 = 0.025 L. Converting 0.025 L gives 25 mL. The remaining volume is made up by adding solvent.
At STP, what volume is occupied by 0.75 mol of an ideal gas?
Correct Answer: D. 16.8 L
Option D is correct because V = n × 22.4 L = 0.75 × 22.4 = 16.8 L. This uses molar volume at STP. Gas volume at fixed conditions is directly proportional to moles.
How many molecules are present in 18 g of water (H2O)?
Correct Answer: B. 6.022 × 10^23
Option B is correct because 18 g of water is 1 mole of H2O. One mole contains 6.022 × 10^23 molecules. Water’s molar mass is 18 g/mol, so 18 g corresponds to exactly one mole.
For acid-base reactions, 1.0 M H2SO4 solution is approximately equal to?
Correct Answer: A. 2.0 N
Option A is correct because H2SO4 can provide 2 H+ ions, so n-factor is 2. Normality = molarity × n-factor = 1.0 × 2 = 2.0 N. Normality depends on the reaction type and n-factor.
How many moles are present in 3.011 × 10^23 molecules of a substance?
Correct Answer: C. 0.50 mol
Option C is correct because moles = number of molecules/NA = 3.011 × 10^23 / 6.022 × 10^23. This equals 0.50 mol. Avogadro’s number converts particles to moles.
What is the mass of 0.20 mol of aluminium atoms? (Al = 27 g/mol)?
Correct Answer: A. 5.4 g
Option A is correct because mass = moles × molar mass = 0.20 × 27 = 5.4 g. Molar mass gives grams per mole. This converts chemical amount into mass directly.
For N2 + 3H2 → 2NH3, if 1 mol N2 reacts with 2 mol H2, the maximum NH3 formed is?
Correct Answer: B. 1.33 mol
Option B is correct because 2 mol H2 is limiting since 3 mol H2 is needed for 1 mol N2. Extent = 2/3, so NH3 formed = 2 × (2/3) = 4/3 = 1.33 mol. Limiting reagent decides the maximum product formed.
The density of a gas at STP is 2.0 g/L. Its molar mass is approximately?
Correct Answer: B. 44.8 g/mol
Option B is correct because molar mass ≈ density × molar volume at STP = 2.0 × 22.4. This gives 44.8 g/mol. This relation is used for gases near ideal behavior at STP.
An impure CaCO3 sample of mass 10 g releases 2.2 g CO2 on heating (CaCO3 → CaO + CO2). The percentage purity of CaCO3 is?
Correct Answer: D. 50%
Option D is correct because moles CO2 = 2.2/44 = 0.05 mol, so moles CaCO3 = 0.05 mol. Mass of pure CaCO3 = 0.05 × 100 = 5 g, so purity = (5/10) × 100 = 50%. This uses stoichiometry of 1:1 between CaCO3 and CO2.