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Question: Rearrange the following in the order of increasing mass and choose the correct order (Atomic mass, \...

Rearrange the following in the order of increasing mass and choose the correct order (Atomic mass, N=14,O=16,Cu=32N = 14,O = 16,Cu = 32)
I.1 molecule of oxygen
II.1 atom of nitrogen
III.1×10−10gm1 \times {10^{ - 10}}gm molecule of oxygen
IV.1×10−10gm1 \times {10^{ - 10}}gm of copper
A.II<I<IV<IIIII < I < IV < III
B.IV<III<I<IVIV < III < I < IV
C.II<III<I<IVII < III < I < IV
D.III<IV<I<IIIII < IV < I < II

Explanation

Solution

We can order the elements by increasing mass by calculating the amount of the elements in grams. We can determine the amount of elements in grams by converting the particles which include molecules, atoms into grams. We have to convert the molecules, atoms into grams with the help of Avogadro number and molecular mass. The value of Avogadro number is 6.022×10236.022 \times {10^{23}}.

Complete step by step answer: 1 molecule of oxygen
One molecule of oxygen has two atoms of oxygen. The atomic mass of oxygen is 16g16g. So, the molecular mass of oxygen (O2)\left( {{O_2}} \right) is 32g32g. We can now calculate the mass of oxygen in grams using the Avogadro number.
We can calculate the mass of oxygen in grams as,
Mass of oxygen=326.022×1023\dfrac{{32}}{{6.022 \times {{10}^{23}}}}
Mass of oxygen=5.3×10−23g5.3 \times {10^{ - 23}}g
The mass of 1 molecule of oxygen in grams is 5.3×10−23g5.3 \times {10^{ - 23}}g.
1 atom of nitrogen
In this case, we have to first convert the atoms of nitrogen into moles of nitrogen. The moles of nitrogen are converted into grams using the molar mass of nitrogen.
We can convert the atoms of nitrogen into moles of nitrogen using Avogadro number.
Moles of nitrogen=16.022×1023\dfrac{1}{{6.022 \times {{10}^{23}}}}
Moles of nitrogen=1.66×10−24moles1.66 \times {10^{ - 24}}moles
The moles of nitrogen is 1.66×10−24moles1.66 \times {10^{ - 24}}moles.
We can calculate the mass of nitrogen in grams as,
We know that 14g14g is atomic mass of nitrogen.
Mass of nitrogen=1.66×10−24×141.66 \times {10^{ - 24}} \times 14
Mass of nitrogen=23.2×10−24g23.2 \times {10^{ - 24}}g
The mass of 1 atom of nitrogen in grams is 23.2×10−24g23.2 \times {10^{ - 24}}g.
1×10−10gm1 \times {10^{ - 10}}gm molecule of oxygen
We have to know that 1×10−10gm1 \times {10^{ - 10}}gm molecule of oxygen is equal to 1×10−10gm1 \times {10^{ - 10}}gm mole of oxygen. We can now calculate the mass of 1×10−10gm1 \times {10^{ - 10}}gm molecule of oxygen using the molecular mass of oxygen.
The molecular mass of oxygen is 32g32g.
We can calculate the mass 1×10−10gm1 \times {10^{ - 10}}gm molecule of oxygen as,
Mass of 1×10−10gm1 \times {10^{ - 10}}gm molecule of oxygen=1×10−10×321 \times {10^{ - 10}} \times 32
Mass of 1×10−10gm1 \times {10^{ - 10}}gm molecule of oxygen=3.2×10−9g3.2 \times {10^{ - 9}}g
The mass of 1×10−10gm1 \times {10^{ - 10}}gm molecule of oxygen is 3.2×10−9g3.2 \times {10^{ - 9}}g.
1×10−10gm1 \times {10^{ - 10}}gm of copper
The mass of 1 molecule of oxygen in grams is 5.3×10−23g5.3 \times {10^{ - 23}}g.
The mass of 1 atom of nitrogen in grams is 23.2×10−24g23.2 \times {10^{ - 24}}g.
The mass of 1×10−10gm1 \times {10^{ - 10}}gm molecule of oxygen is 3.2×10−9g3.2 \times {10^{ - 9}}g.
The mass of 1×10−10gm1 \times {10^{ - 10}}gm of copper is 1×10−10gm1 \times {10^{ - 10}}gm.
Let us compare the masses of all four, we get (2)<(1)<(4)<(3)\left( 2 \right) < \left( 1 \right) < \left( 4 \right) < \left( 3 \right). So let us now arrange according to the increasing order as (II)<I<(IV)<(III)\left( {II} \right) < I < \left( {IV} \right) < \left( {III} \right).
Therefore, the option (A) is correct.

Note: We have to remember that atomic mass in units is converted to grams by multiplying the Avogadro number. We have to know that different molecules contain different weights. Thus, one mole of molecule (or) atom is different in gram also. If we don’t convert the units to required units there would be an error in the final answer.