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Question: One mole of any substance contains \(6.022 \times {10^{23}}\) atoms/molecules. Number of molecules o...

One mole of any substance contains 6.022×10236.022 \times {10^{23}} atoms/molecules. Number of molecules of H2SO4{H_2}S{O_4} present in 100mL of 0.02M H2SO4{H_2}S{O_4} ​ solution is __________?
a.) 12.044×102012.044 \times {10^{20}} molecules.
b.) 6.022×10236.022 \times {10^{23}} molecules.
c.) 1×10231 \times {10^{23}} molecules.
d.) 12.044×102312.044 \times {10^{23}} molecules.

Explanation

Solution

Start by using the definition and formula of Molarity , defined as number of moles present per litre of volume of solution. Rearrange the terms and find out the number of moles present in 100mL of 0.02M H2SO4{H_2}S{O_4} ​ solution by converting into desired units. Then find out the total number of molecules present in the same by multiplying Avogadro’s number or constant NA=6.022×1023{N_A} = 6.022 \times {10^{23}}.

Complete step by step answer:
Given,
100mL of 0.02M H2SO4{H_2}S{O_4}
1 mole = 6.022×10236.022 \times {10^{23}} atoms/molecule………….(equation 1)
6.022×10236.022 \times {10^{23}} is nothing but the Avogadro’s number denoted by NA{N_A}.

Now , we know that
Molarity (M) = No. of moles(n)Volume of solution in Litres(V)\dfrac{{{\text{No}}{\text{. of moles(n)}}}}{{{\text{Volume of solution in Litres(V)}}}}
Using the above formula we can find the number of moles present in 100mL of 0.02M H2SO4{H_2}S{O_4}
solution
No. of moles(n)=Volume of solution in Litres(V)×Molarity(M){\text{No}}{\text{. of moles(n)}} = {\text{Volume of solution in Litres(V}}) \times {\text{Molarity(M)}}
But before that let us convert 100 ml into Litres , which will be equal to 1001000=0.1L\dfrac{{100}}{{1000}} = 0.1L
Now substituting all the values in above discussed formula , we get
No. of moles(n)=0.1×0.02 No. of moles(n)=2×10−3moles  {\text{No}}{\text{. of moles(n)}} = 0.1 \times 0.02 \\\ {\text{No}}{\text{. of moles(n)}} = 2 \times {10^{ - 3}}{\text{moles}} \\\
Now from equation 1 , we get
1 mole =NA{N_A}= 6.022×10236.022 \times {10^{23}} atoms/molecule
So 2×10−3moles2 \times {10^{ - 3}}{\text{moles}} = (2×10−3)×(6.022×1023)(2 \times {10^{ - 3}}) \times (6.022 \times {10^{23}})molecules
=12.044×1020= 12.044 \times {10^{20}}molecules.

Which means 100mL of 0.02M H2SO4{H_2}S{O_4} solution contains 12.044×102012.044 \times {10^{20}}molecules.
So, the correct answer is “Option A”.

Note: Students must know all the important laws and definitions related to mole concept which includes Molarity , Molality , Normality. Also one must remember Avogadro’s number(NA{N_A}) i.e. 6.022×10236.022 \times {10^{23}}, which is required to solve many similar questions. Attention must be given while substituting the values and the exponents, as any mistake might lead to wrong or absurd value.