Question

Find the number of molecules present in 70 g dinitrogen.

• A. \(1.5055\times 10^{24}\)
• B. \(1.5055\times 10^{23}\)
• C. \(3.011\times 10^{24}\)
• D. \(3.011\times 10^{23}\)

Hint:

Convert 70 g to moles using the molar mass 28 g/mol, then multiply by Avogadro's number.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The number of molecules in a sample equals the number of moles times Avogadro's number. So we first convert the given mass of dinitrogen into moles.
Dinitrogen is the diatomic molecule \(\text{N}_2\).

Step 2: Key Formula or Approach:
1. Molar mass of \(\text{N}_2 = 2 \times 14 = 28 \text{ g mol}^{-1}\).
2. Moles \(n = \dfrac{\text{mass}}{\text{molar mass}}\).
3. Number of molecules \(= n \times N_A\), where \(N_A = 6.022\times 10^{23} \text{ mol}^{-1}\).

Step 3: Detailed Explanation:
Moles of dinitrogen:
\[ n = \frac{70}{28} = 2.5 \text{ mol} \]
Number of molecules:
\[ N = 2.5 \times 6.022\times 10^{23} = 15.055\times 10^{23} = 1.5055\times 10^{24} \]
Option (B) is ten times too small, which would happen if the moles were taken as 0.25. Option (C) would need 5 mol (140 g) and option (D) would need 0.5 mol (14 g), so neither matches 70 g.

Final Answer:
70 g of \(\text{N}_2\) is 2.5 mol, which holds \(1.5055\times 10^{24}\) molecules. This is option (A). \[ \boxed{1.5055\times 10^{24}\text{ molecules (A)}} \]