 Amount of oxygen in ${ 32.2g }$ of ${ Na }_{ 2 }{ SO }_{ 4 }{ .10H }_{ 2 }{ O }$ is:A.${ 20.8g }$B.${ 22.4g }$C.${ 2.24g }$D.${ 2.08g }$ Verified
147.9k+ views
Hint:The gram molecule is defined as the number of molecules present in one mole of that substance instead of atoms.
Gram molecular weight:- It is defined as the mass of one mole of a compound equal in grams to the molecular weight.

It is given that:
The given compound is ${ Na }_{ 2 }{ SO }_{ 4 }{ .10H }_{ 2 }{ O }$
Molar mass of ${ Na }_{ 2 }{ SO }_{ 4 }{ .10H }_{ 2 }{ O } = { 46+32+64+180 } = { 322g/mol }$
The given weight of the compound ${ Na }_{ 2 }{ SO }_{ 4 }{ .10H }_{ 2 }{ O } = { 32.2g }$
Therefore, the number of moles = $\dfrac{Given mass}{Molar mass}$
= $\dfrac{32.2g}{322gmol^{-1}}$
So, number of moles = ${ 0.1 }$ moles
Now, number of moles of oxygen in ${ Na }_{ 2 }{ SO }_{ 4 }{ .10H }_{ 2 }{ O } = { 4+10 } = { 14 }$
Therefore, the number of moles of oxygen = ${ 0.1\times 14\quad =1.4moles }$
Now, we can calculate the weight of oxygen = ${ no.\quad of\quad moles\times molecular\quad weight }$
= ${ 1.4moles\times 16gmol }^{ -1 } = { 22.4g }$
Hence, the amount of oxygen in ${ 32.2g }$ of ${ Na }_{ 2 }{ SO }_{ 4 }{ .10H }_{ 2 }{ O }$ is ${ 22.4g }$.

The correct option is B.

Note:
The formula of the mole concept can be written as:
-The number of atoms or molecules = Avogadro’s no.
-The relationship between amu and gram is; ${ 1amu } = { 1 } g / { N }_{ a } = { 1.66\times { 10 }^{ -24 } }$, where N$_a$ = Avogadro’s number.
-Therefore, the mass of one mole of an element is equal to its atomic mass in grams.
-The number of moles = Mass of the sample/molar mass.