Question

The largest no. of molecules are in:
(A) $36\,g\,{H_2}O$
(B) $28\,g\,CO$
(C) $46\,g\,{C_2}{H_5}$
(D) $54\,g\,{N_2}{O_5}$

Answer 1

As we know that the compounds have their own molar mass which can be calculated by the atomic masses of the atoms in that compound. According to Avogadro's law, 1 mole of the compound contains $6.023\,\times\,{10^{23}}\,$ atoms or molecules or ions.

Complete step by step answer:
According to the mole concept 1 mole of the compound represents $6.023\,\times\,{10^{23}}\,$ particles. The number of $6.023\,\times\,{10^{23}}\,$ is called Avogadro number and is represented as ${N_A}$.
So, the number of molecules can be calculated if we calculate the number of moles of the given compounds. For this, we will use a formula as-
$Number\,of\,moles\,\left( n \right) = {\dfrac{{Given\,mass\left( g \right)}}{Molar\,mass}} - - - - \left( i \right)$
Let’s calculate the number of moles of option (A).
The number of moles of ${H_2}O$ can be obtained by putting values in equation $(i)$
Given mass of {H_2}O$$$$ = 36\,g
and
Molar mass of ${H_2}O = 2\,\times\,1\,\,g + \,16\,g = 18\,g$ where atomic mass of hydrogen $= 1\,g$, atomic mass of oxygen $= 16\,g$
The number of moles of {H_2}O$$$$ = \dfrac{{36\,g}}{{18}} = 2\,mole
As we know that $1\,$ mole $= 6.023\,\times\,{10^{23}}\,$ molecules of water
Therefore, 2 mole $= 2\,\times\,6.023\,\times\,{10^{23}}\, = 12.05\,\times\,{10^{23}}\,$ molecules of water
To calculate the number of moles of option (B).
The number of moles of $CO$ can be obtained by putting values in equation $(i)$
Given mass of $CO = 28 g$
and
Molar mass of $CO = 1\,\times\,12\,\,g + \,16\,g = 28\,g$ where atomic mass of carbon$= 12g$, atomic mass of oxygen = 16 g
The number of moles of CO$$$$ = \dfrac{{28\,g}}{{28}} = 1\,mole
As we know that 1mole $= 6.023\,\times\,{10^{23}}\,$ molecules of $CO$
To calculate the number of moles of option (C)
The number of moles of ${C_2}{H_5}$ can be obtained by putting values in equation $(i)$
Given mass of ${C_2}{H_5}$ = 46 g
and
Molar mass of ${C_2}{H_5} = 2\,\times\,12\,\,g + \,1\,\times\,5\,g = 29\,g$ where atomic mass of carbon = 12g, atomic mass of hydrogen = 1g
The number of moles of {C_2}{H_5}$$$$ = \dfrac{{46\,g}}{{29}} = 1.59\,mole
As we know that $1\,$mole $= 6.023\,\times\,{10^{23}}\,$ molecules of ${C_2}{H_5}$
Therefore, 1.59 mole $= 1.59\,\times\,6.023\,\times\,{10^{23}}\, = 9.55\,\times\,{10^{23}}\,$ molecules of ${C_2}{H_5}$.
To calculate the number of moles of option (D)
The number of moles of ${N_2}{O_5}$ can be obtained by putting values in equation $(i)$
Given mass of ${N_2}{O_5}$ = 54 g
and
Molar mass of ${N_2}{O_5} = 2\,\times\,14\,\,g + \,5\,\times\,16\,g = 108\,g$ where atomic mass of nitrogen = 14g, atomic mass of oxygen = 16 g
The number of moles of {N_2}{O_5}$$$$ = \dfrac{{54\,g}}{{108\,}} = 0.5\,mole
As we know that 1 mole $= 6.023\,\times\,{10^{23}}\,$ molecules of ${N_2}{O_5}$
Therefore, 0.5 mole $= 0.5\,\times\,6.023\,\times\,{10^{23}}\, = 3.011\,\times\,{10^{23}}\,$ molecules of ${N_2}{O_5}$.

Hence, the correct option is (A).

Note: As the molar mass of the compounds increases, the number of molecules present in that compound decreases. Here the maximum number of molecules are present in 36 g of $H_2O$ don’t confuse it with the 28 g $CO$.