<span>Let's </span>assume that water vapor has ideal gas
behavior. <span>
Then we can use ideal gas formula,
PV = nRT<span>
</span><span>Where, P is the pressure of the gas (Pa), V
is the volume of the gas (m³), n is the number
of moles of gas (mol), R is the universal gas constant ( 8.314 J mol</span></span>⁻¹ K⁻¹) and T is temperature in Kelvin.<span>
<span>
</span>P = 1 atm = 101325 Pa (standard pressure)
V = 13.97 L = 13.97 x 10</span>⁻³ m³<span>
n = ?
R = 8.314 J mol</span>⁻¹ K⁻¹<span>
T = 0 °C = 273 K (standard temperature)
<span>
By substitution,
</span>101325 Pa x 13.97x 10</span>⁻³
m³ = n x 8.314 J mol⁻¹ K⁻¹ x 273 K<span>
n = 0.624 mol
<span>
Hence, the moles of water vapor at STP is 0.624 mol.
According to the </span></span>Avogadro's constant, 1 mole of substance has 6.022 × 10²³ particles.
<span>
Hence, number of atoms in water vapor = 0.624 mol x </span>6.022 × 10²³ mol⁻¹
<span> = 3.758 x 10</span>²³<span>
</span>
Answer:
D. (16.0 g + 16.0 g) × 100% / (32.1 g + 16.0 g + 16.0 g) = 49.9%
Explanation:
Step 1: Detemine the mass of O in SO₂
There are 2 atoms of O in 1 molecule of SO₂. Then,
m(O) = 2 × 16.0 g = 16.0 g + 16.0 g = 32.0 g
Step 2: Determine the mass of SO₂
m(SO₂) = 1 × mS + 2 × mO = 1 × 32.1 g + 2 × 16.0 g = 32.1 g + 16.0 g + 16.0 g = 64.1 g
Step 3: Detemine the mass percent of oxygen in SO₂
We will use the following expression.
m(O)/m(SO₂) × 100%
(16.0 g + 16.0 g) × 100% / (32.1 g + 16.0 g + 16.0 g) = 49.9%
