Question

How many molecules are in 13.5 g of sulfur dioxide?

Answer 1

oxygen-8×2=16, sulphur=16
1mole=6.023×10^23molecule=22.4l=32gram

32g=6.023×10^23molecule
1g=6.023×10^23÷32
13.5g=6.023×10^23÷32×13.5
=2.5×10^23 molecule
.°. there are 2.5×10^23 molecule in 13.5g of sulphur dioxide

Answer 2

Sulfur dioxide is formed by the reaction of sulfur with oxygen. One mole of a substance is equal to 6.022 x 10²³.

In 13.5 grams of sulfur dioxide, there are 1.27 x 10²³ molecules present.

Given that,

• Given mass of sulfur dioxide = 13.5 gram
• Molar mass of sulfur dioxide = 64 g/mol
• $\text{Number of moles} &= \dfrac{\text {given mass}}{\text {molar mass}}$
• $\text{Number of moles} &= \dfrac{13.5\; \text g}{\text {64 g/mol}}$
• Number of moles = 0.21 mol

Now, calculating the amount of molecules in 13.5 grams of sulfur dioxide:

Molecules of SO₂ = 0.21 x 6.022 x 10²³ mol⁻¹.

Molecules of SO₂ = 1.27  x 10²³

Therefore, the molecules in 13.5 g of sulfur dioxide is 1.27  x 10²³.

To know more about Avogadro's number, refer to the following link:

https://brainly.com/question/10010335