Question:
Arrange the following in the order of their increasing masses. \(${(Atomic\, mass : O = 16, N = 14)}$\) I. One atom of oxygen II. One atom of nitrogen III. \($1 \times 10^{-10}$\) mole of oxygen
Arrange the following in the order of their increasing masses. \(${(Atomic\, mass : O = 16, N = 14)}$\) I. One atom of oxygen II. One atom of nitrogen III. \($1 \times 10^{-10}$\) mole of oxygen
Updated On: Jun 23, 2023
- II < I < III
- I < II < III
- III < II < I
- II < III < I
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The Correct Option is A
Solution and Explanation
I. Mass of one atom of oxygen
$= \frac{16}{6.023\times10^{23}} = 2.66 \times10^{-23} g$
II. Mass of one atom of nitrogen
$= \frac{14}{6.023\times10^{23}} = 2.32 \times10^{-23} g$
III. Mass of $1 \times 10^{-10}$ mole of oxygen
$= 16 \times 10^{-10} g$
So, the order of increasing masses is $II < I < III$.
$= \frac{16}{6.023\times10^{23}} = 2.66 \times10^{-23} g$
II. Mass of one atom of nitrogen
$= \frac{14}{6.023\times10^{23}} = 2.32 \times10^{-23} g$
III. Mass of $1 \times 10^{-10}$ mole of oxygen
$= 16 \times 10^{-10} g$
So, the order of increasing masses is $II < I < III$.
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