Question

How many moles of $Mg{S_2}{O_3}$ are in $241\;{\text{g}}$ of the compound?

Answer 1

Hint: This is a very easy problem which can be solved with the help of basic fundamentals of mole concept. We must remember what a mole is in order to solve this question. And later, we will use basic unitary methods to find the required number of moles.

Formula Used:
\[{\text{mole}} = \dfrac{{{\text{mass}}}}{{{\text{molar mass}}}}\]

Complete step-by-step answer:According to the question, the weight of $Mg{S_2}{O_3}$ provided to us is given as $241\;{\text{g}}$
The molecular mass of $Mg{S_2}{O_3}$ is \[136.43 {\text{g mol}}{{\text{ }}^{{\text{ - 1}}}}\]
Also, we know that the molecular weight of any compound has exactly one mole of entities.
According to the above statement,
\[136.43 g\]of $Mg{S_2}{O_3}$ has \[1\] mole of $Mg{S_2}{O_3}$
So, $241\;{\text{g}}$ of $Mg{S_2}{O_3}$ will have \[\dfrac{1}{{136.43 {\text{g mo}}{{\text{l}}^{{\text{ - 1}}}}}} \times 241 {\text{g}}\] moles of the compound
Upon solving, we get number of moles of \[Mg{S_2}{O_3} = 1.77\]
Hence, $241\;{\text{g}}$ of $Mg{S_2}{O_3}$ has \[1.77\] moles of the compound

Additional Information:
The number \[6.022 \times {10^{23}}\] is commonly referred to as the constant of Avogadro and is often denoted by the '\[{N_A}\]' symbol. Atoms, molecules, monatomic/polyatomic ions, and other particles can be the fundamental entities (such as electrons) that can be represented in moles.
A mole is defined in the field of chemistry as the quantity of a substance containing exactly \[6.022 \times {10^{23}}\] 'elementary entities' of the substance given.

Note:Even one gram of a pure element is known to contain an enormous number of atoms when dealing with particles at an atomic (or molecular) level. This is where it is widely used to use the mole concept. It focuses primarily on the unit known as a mole, which is a very large number of particles counted.