Question

If we assume that one-sixth the mass of an atom of ${}^{12}C$ isotope is taken as the reference, the mass of one molecule of oxygen will:

• (A) Be double its original value
• (B) Be half its original value
• (C) Be the same
• (D) Increase by four fold

Answer 1

The Correct Option is C

Explanation:
If we assume that one-sixth the mass of an atom of ${}^{12}C$ isotope is taken as the reference, the mass of one molecule of oxygen will be the same.We know that: $12\mathrm{g}$ of Carbon $=6.023\times {10}^{23}$ atomsSo, $1g=\frac{6.023\times {10}^{23}}{12}$$\therefore 6g$ of carbon $=6\times \frac{6.023\times {10}^{23}}{12}$$=\frac{1}{2}\times 6.023\times {10}^{23}$This is our new Avogadro's number according to the given conditions.Then, 1 atomic mass unit $=\frac{1}{N_A}$$=\frac{2}{6.023\times {10}^{23}}$Mass of 1 mole of $O_2=32\times$ Avogrado's number $\times \frac{2}{6.023\times {10}^{23}}$$=32\times \frac{6.023\times {10}^{23}}{2}\times \frac{2}{6.023\times {10}^{23}}$$=32g$Therefore, there is no change in the mass number of $O_2$.Hence, the correct option is (C).

Answer 2

Given:
One-sixth of the mass of a \(^{12}C\) atom is taken as the reference mass.

Reference mass:
Let \(m_{^{12}C}\) be the mass of one \(^{12}C\) atom.
Reference mass \(= \frac{1}{6} \times m_{^{12}C}\).

Mass of oxygen molecule O₂:
The oxygen molecule O₂ consists of 2 oxygen atoms.
Let \(m_{^{16}O}\) be the mass of one \(^{16}O\) atom.

The total mass of the O₂ molecule:
 The total mass of the O₂ molecule \(= 2 \times m_{^{16}O}\).

Since the reference mass is based on one-sixth of the mass of a \(^{12}C\) atom and does not affect the actual mass of oxygen atoms (assuming \(^{16}O\) isotopes), the mass of one molecule of oxygen O₂ relative to this reference will be the same.

So, the correct answer is (C): be the same