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/1M
A sample of gold: xM = 35.12g/1000ml ⇒ 1000 ml = 35.12g/x
1000 ml = 196.97g/1M = 35.12g/x
⇒ 196.97g/1M = 35.12g/x
x = 35.12g / 196.97g * 1M = 0.1783M
Q1. 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 mole
From 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 M
After crossing the products:
x = 6.02× 10²³atoms * 0.1783 M / 1M = 1.073 × 10²³ atoms
Q2. 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 g
A mole (M) is an atomic or molar mass in 1000 ml:
Sucrose: 1M = 342g/1000ml ⇒ 1000 ml = 342g/1M
A sample of sucrose: xM = 1.202g/1000ml ⇒ 1000 ml = 1.202g/x
1000 ml = 342g/1M = 1.202g/x
⇒ 342g/1M = 1.202g/x
x = 1.202g / 342g * 1M = 0.0035 M
Q2. b) The answers are:
- carbon: 0.042 moles
- hydrogen: 0.077 moles
- oxygen: 0.0385 moles
In a sample of sucrose of 0.0035 M, there are 12 atoms of carbon:
12 * 0.0035M = 0.042 M
In a sample of sucrose of 0.0035 M, there are 22 atoms of hydrogen:
22 * 0.0035M = 0.077 M
In a sample of sucrose of 0.0035 M, there are 11 atoms of oxygen:
11 * 0.0035M = 0.0385 M
Q2. c) The answers are:
- carbon: 2.5 × 10²⁴ atoms
- hydrogen: 4.6 × 10²⁴ atoms
- oxygen: 2.3 × 10²⁴ 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 mole
- carbon: 0.042 moles (from the previous task)
Now, let's make a proportion:
6.02 × 10²³atoms : 1M = x : 0.042 M
After 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 M
After 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 M
After crossing the products:
x = 6.02× 10²³atoms * 0.0385 M / 1M = 0.23 × 10²³ atoms = 2.3 × 10²⁴ atoms