Question

How many oxygen atoms are present in 0.36 g of a drop of water at STP?

• A. $6.023 \times 10^{22}$
• B. $1.205 \times 10^{22}$
• C. $6.023 \times 10^{23}$
• D. $1.205 \times 10^{23}$

Hint:

Using a simple frame or just bolding for the box Key Points: Molar mass of H$_2$O $\approx$ 18 g/mol. Avogadro's number (N$_A$) $\approx 6.022 \times 10^{23$ particles/mol. Moles = Mass / Molar Mass. Number of particles = Moles $\times$ N$_A$. Determine the number of atoms of interest per molecule. STP conditions usually apply to gas volume, not mass calculations for liquids/solids.

Answer 1

The Correct Option is B

To determine the number of oxygen atoms in 0.36 g of water (H₂O), we disregard "at STP" since it's irrelevant for a liquid. (A) Calculate the molar mass of water (H₂O):

Molar Mass = 2 \(\times\) (Atomic mass of H) + 1 \(\times\) (Atomic mass of O)
Molar Mass \(\approx\) 2 \(\times\) (1.008 g/mol) + 1 \(\times\) (16.00 g/mol)
Molar Mass \(\approx\) 18.016 g/mol. Using 18 g/mol for simplicity.\n (B) Calculate the number of moles of water:

Moles = \(\frac{\text{Mass}}{\text{Molar Mass}}\)
Moles = \(\frac{0.36 \text{ g}}{18 \text{ g/mol}}\) = 0.02 mol\n (C) Calculate the number of water molecules:

Number of molecules = Moles \(\times\) Avogadro's number (N_A)
Number of molecules = 0.02 mol \(\times\) (6.022 \(\times\) 10²³ molecules/mol)
Number of molecules = \(1.2044 \times 10^{22}\) molecules\n (D) Calculate the number of oxygen atoms:

Each water molecule (H₂O) has one oxygen atom.
Number of oxygen atoms = Number of water molecules \(\times\) (1 oxygen atom / molecule)
Number of oxygen atoms = \(1.2044 \times 10^{22}\) atoms\n Rounding yields \(1.205 \times 10^{22}\) oxygen atoms, which aligns with option (B).