Q1. a) the answer is 0.1783 moles.atomic mass (ar) of gold is 196.97 g. a mole (m) is an atomic or molar mass in 1000 ml: au: 1m = 196.97g/1000ml ⇒ 1000 ml = 196.97g/1ma sample of gold: xm = 35.12g/1000ml ⇒ 1000 ml = 35.12g/x1000 ml = 196.97g/1m = 35.12g/x⇒ 196.97g/1m = 35.12g/x x = 35.12g / 196.97g * 1m = 0.1783mq1. b) the answer is 1.073 × 10²³ atoms.to calculate this, we will use avogadro's number which is the number of units (atoms, molecules) in 1 mole of substance: 6.02 × 10²³ units per 1 molefrom the previous task, we know that the sample of gold has 0.1783 moles.now, let's make a proportion: 6.02 × 10²³atoms : 1m = x : 0.1783 mafter crossing the products: x = 6.02× 10²³atoms * 0.1783 m / 1m = 1.073 × 10²³ atomsq2. a) the answer is 0.0035 moles.let's first calculate molar mass (mr) of sucrose which is a sum of atomic masses (ar) of elements: mr(c₁₂h₂₂o₁₁) = 12ar(c) + 22ar(h) + 11ar(o) = 12*12 + 22*1 + 11*16 = = 144 + 22 + 176 = 342 ga mole (m) is an atomic or molar mass in 1000 ml: sucrose: 1m = 342g/1000ml ⇒ 1000 ml = 342g/1ma sample of sucrose: xm = 1.202g/1000ml ⇒ 1000 ml = 1.202g/x1000 ml = 342g/1m = 1.202g/x⇒ 342g/1m = 1.202g/x x = 1.202g / 342g * 1m = 0.0035 mq2. b) the answers are: - carbon: 0.042 moles- hydrogen: 0.077 moles- oxygen: 0.0385 molesin a sample of sucrose of 0.0035 m, there are 12 atoms of carbon: 12 * 0.0035m = 0.042 min a sample of sucrose of 0.0035 m, there are 22 atoms of hydrogen: 22 * 0.0035m = 0.077 min a sample of sucrose of 0.0035 m, there are 11 atoms of oxygen: 11 * 0.0035m = 0.0385 mq2. c) the answers are: - carbon: 2.5 × 10²⁴ atoms- hydrogen: 4.6 × 10²⁴ atoms- oxygen: 2.3 × 10²⁴ atomsto calculate this, we will use avogadro's number which is the number of units (atoms, molecules) in 1 mole of substance: 6.02 × 10²³ units per 1 mole- carbon: 0.042 moles (from the previous task)now, let's make a proportion: 6.02 × 10²³atoms : 1m = x : 0.042 mafter crossing the products: x = 6.02× 10²³atoms * 0.042 m / 1m = 0.25 × 10²³ atoms = 2.5 × 10²⁴ atoms- hydrogen: 0.077 moles (from the previous task)now, let's make a proportion: 6.02 × 10²³atoms : 1m = x : 0.077 mafter crossing the products: x = 6.02× 10²³atoms * 0.077 m / 1m = 0.46 × 10²³ atoms = 4.6 × 10²⁴ atoms- oxygen: 0.0385 moles (from the previous task)now, let's make a proportion: 6.02 × 10²³atoms : 1m = x : 0.0385 mafter crossing the products: x = 6.02× 10²³atoms * 0.0385 m / 1m = 0.23 × 10²³ atoms = 2.3 × 10²⁴ atoms read more on -