Question

‌Calculate‌ ‌the‌ ‌number‌ ‌of‌ ‌moles‌ ‌in‌ ‌\[52g\] of‌ ‌\[He\]‌ ‌(finding‌ ‌mole‌ ‌from‌ ‌mass).‌ ‌

Answer 1

Hint: This type of question is directly formula-based. Simply use the formula of the number of moles and put the given data and calculate the number of moles of \[He\] asked in the question. Mass of \[He\] and molar mass is known from the question.

Complete step-by-step answer: Given data:
Mass of \[He\]: \[52g\]
The molar mass of \[He\]: \[gmo{l^{ - 1}}\]
Using the given data in the number of moles formula we will get:
Number of moles \[(n) = \dfrac{{{W_{He}}}}{{{M_{He}}}}\dfrac{{52}}{4} = 13mol\]
So the number of moles in \[52g\] of \[He\] is \[13mol\]
Additional information: Relation between mole, mass and number of atoms:
Let’s refresh our memory of what a mole is. Just as a dozen is to 12 things a mole is to \[6.022 \times {10^{23}}\] the number of anything if I had given you 4 dozen of sulphur you would have multiplied 4 by 12 to find the number of sulphur atoms. Now I have \[4mol\] of sulphur. How many atoms do you think that there are in this sample? Since I am talking about \[mole\] you can simply multiply 4 with \[6.022 \times {10^{23}}\]. We will get:
\[
  4 \times 6.022 \times {10^{23}} \\
   = 24.088 \times {10^{23}} \\
 \]
 It’s a pretty huge number that this sample contains. So in case, we have \[5mol\] of sulphur to calculate the number of atoms present we will multiply \[5mol\] with \[6.022 \times {10^{23}}\] which gives us \[3.011 \times {10^{23}}\] the number of sulphur atoms.
In general, if we have n moles then the number of atoms will be \[n \times 6.022 \times {10^{23}}\]or \[n \times {N_A}\]
Here \[{N_A}\] is Avogadro's number

Note: Molar mass is equal to the sum of number of protons and number of neutrons. In \[He\], there are two protons and two neutrons. That is why the molar mass of \[He\] is \[4amu\]. We use unitary methods to solve these questions. Relationship between number of atoms and number of moles is given by:
Number of atoms= number of \[moles \times {N_A}\]