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

Hint: 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 $CO$.