Arrange the following in the order of increasing mass (atomic mass : O = 16, Cu = 63, N = 14)(I) One atom of oxygen (ll) One atom of nitrogen(III) 1x 10-10 mole of oxygen (lV) 1x 10-10 mole of copper

A
II < l < lll < lV
B
I < ll < lll < lV
C
III < ll < lV < l
D
IV < ll < lll < l

Solution:

Mass of one atom of oxygen = $\frac{16}{6.023 \times 10^{23}} = 2.66 \times 10^{- 23}$
Mass of one atom of nitrogen = $\frac{14}{6.023 \times 10^{23}} = 2.32 \times 10^{- 23}$
$Mass of 1 \times 10^{- 10} mole of oxygen = 16 \times 10^{- 10} Mass of 1 \times 10^{- 10} mole of copper = 63 \times 1 \times 10^{- 10} = 63 x 10^{- 10}$
So, the order of increasing mass is II < I < III < IV.

Arrange the following in the order of increasing mass (atomic mass : O = 16, Cu = 63, N = 14)
(I) One atom of oxygen
(ll) One atom of nitrogen
(III) 1x 10^-10 mole of oxygen
(lV) 1x 10^-10 mole of copper

Solution:

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