Laws of chem combination, mole concept, formulae MCQs

Q1. How many grams of pure NaOH (M = 40) are required to prepare 200 mL of a 0.5 mol/L solution?

• A.4.0 g
• B.2.0 g
• C.8.0 g
• D.20.0 g

Answer: A. 4.0 g

Start by listing the data - 200 mL, 0.5 mol (chemistry, chapter 'Atoms and Molecules'). What we must find: How many grams of pure NaOH (M = 40) are required to prepare 200 mL of a 0. From the chapter we use the relation: Moles required = 0.5 × 0.200 = 0.10 mol. The arithmetic is: Moles required = 0.5 × 0.200 = 0.10 mol → Mass = 0.10 × 40 = 4.0 g. Matching this against the options, A) 4.0 g is the answer.

Q2. Equal masses (4 g each) of H2, He, O2 and SO2 are taken. Which sample contains the greatest number of molecules?

• A.H2
• B.He
• C.O2
• D.SO2

Answer: A. H2

We are told: 4 g. To find: the unknown asked in the stem. Formula - Moles in 4 g: H2 = 4/2 = 2. This is the equation that links the given quantities to the unknown (chemistry, chapter 'Atoms and Molecules'). Substituting: Moles in 4 g: H2 = 4/2 = 2 → He = 4/4 = 1 → O2 = 4/32 = 0.125 → SO2 = 4/64 = 0.0625. Why this is the right approach - H2 has the most moles, so the most molecules. Putting it together the answer is A) H2.

Q3. What mass of oxygen is required to burn 6.0 g of carbon completely to CO2?

• A.16 g
• B.8 g
• C.32 g
• D.12 g

Answer: A. 16 g

Spot-the-setup - a typical chemistry numerical. The data on the table: 6.0 g. We are after the quantity the stem asks for. Tool of choice - So 6 g C reacts with (32/12) × 6 = 16 g O2. Rearrange it for the unknown before substituting. Numbers in: So 6 g C reacts with (32/12) × 6 = 16 g O2. Common-sense check: C + O2 → CO2: 12 g C reacts with 32 g O2. Lock in option A) 16 g.