Question

particles to volume and mass

1. what volume (in liters) does 1.8066 × 10²⁴ particles of methane gas (ch₄) occupy at stp?

(avogadro’s number = 6.022 × 10²³ particles/mol)

1. find the mass (in grams) of 3.011 × 10²³ particles of neon gas (ne).

(molar mass of ne = 20.18 g/mol; avogadro’s number = 6.022 × 10²³ particles/mol)

1. how many particles are in 11.2 l of argon gas (ar) at stp?

(avogadro’s number = 6.022 × 10²³ particles/mol)

Explanation:

Let's solve question 8: Find the mass (in grams) of \( 3.011 \times 10^{23} \) particles of neon gas (Ne). (Molar mass of Ne = 20.18 g/mol; Avogadro’s number = \( 6.022 \times 10^{23} \) particles/mol)

Step 1: Find moles of Ne

Moles (\( n \)) = \( \frac{\text{Number of particles}}{\text{Avogadro's number}} \)
\( n = \frac{3.011 \times 10^{23}}{6.022 \times 10^{23}} \)
\( n = 0.5 \) mol

Step 2: Find mass of Ne

Mass (\( m \)) = Moles (\( n \)) × Molar mass (\( M \))
\( m = 0.5 \, \text{mol} \times 20.18 \, \text{g/mol} \)
\( m = 10.09 \, \text{g} \)

Answer:

The mass of \( 3.011 \times 10^{23} \) particles of neon gas is \( \boldsymbol{10.09} \) grams.