
The unit of molar susceptibility is
\[\begin{align}
& A)k{{g}^{-1}}{{m}^{3}} \\
& B)kg{{m}^{3}} \\
& C){{m}^{3}} \\
& D)unitless \\
\end{align}\]
Answer
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Hint: We will find the unit of molar susceptibility from the expression to find molar susceptibility. We will assign the units for the parameters in the expression and finally deduce the unit by cancellation of similar units. We must keep in mind that volume susceptibility is a unit less quantity.
Formula used: \[{{\chi }_{m}}=\dfrac{M{{\chi }_{v}}}{\rho }\]
Complete answer:
Firstly we must understand molar susceptibility. Molar susceptibility is a measure of magnetic susceptibility which is given as the ratio of product of molar mass \[M\] and volume susceptibility \[{{\chi }_{v}}\] to density \[\rho \]of the material.
Molar mass \[M\] is measured in \[kg/mol\]
Now, volume susceptibility \[{{\chi }_{v}}\], is a dimensionless proportionality constant that indicates the degree of magnetization of a material in response to the magnetic field. It is determined by the formula,
\[\begin{align}
& M={{\chi }_{v}}H \\
& {{\chi }_{v}}=\dfrac{M}{H} \\
\end{align}\]
Where, \[M\] is the magnetization of the material which is measured in \[A/m\] and \[H\] is the magnetic field strength which is also measured in \[A/m\].
Therefore, we can understand that volume susceptibility \[{{\chi }_{v}}\] is a dimensionless quantity.
Then, \[\rho \] is the density of the given material which has the usual unit of \[kg/{{m}^{3}}\].
If we substitute all these units to the expression of molar susceptibility,
\[\begin{align}
& {{\chi }_{m}}=\dfrac{M{{\chi }_{v}}}{\rho } \\
& \Rightarrow {{\chi }_{m}}=\dfrac{\dfrac{kg}{mol}}{\dfrac{kg}{{{m}^{3}}}}={{m}^{3}}/mol \\
\end{align}\]
So we got the unit of molar susceptibility as \[{{m}^{3}}/mol\] but it is actually represented as \[{{m}^{3}}\].
Therefore the correct answer is option c.
Note: Molar magnetic susceptibility is also used to classify between paramagnetic and diamagnetic materials. We must always keep in mind that volume susceptibility is a dimensionless quantity. Susceptibility can also be measured as mass susceptibility which has the unit of \[{{m}^{3}}/kg\].
Formula used: \[{{\chi }_{m}}=\dfrac{M{{\chi }_{v}}}{\rho }\]
Complete answer:
Firstly we must understand molar susceptibility. Molar susceptibility is a measure of magnetic susceptibility which is given as the ratio of product of molar mass \[M\] and volume susceptibility \[{{\chi }_{v}}\] to density \[\rho \]of the material.
Molar mass \[M\] is measured in \[kg/mol\]
Now, volume susceptibility \[{{\chi }_{v}}\], is a dimensionless proportionality constant that indicates the degree of magnetization of a material in response to the magnetic field. It is determined by the formula,
\[\begin{align}
& M={{\chi }_{v}}H \\
& {{\chi }_{v}}=\dfrac{M}{H} \\
\end{align}\]
Where, \[M\] is the magnetization of the material which is measured in \[A/m\] and \[H\] is the magnetic field strength which is also measured in \[A/m\].
Therefore, we can understand that volume susceptibility \[{{\chi }_{v}}\] is a dimensionless quantity.
Then, \[\rho \] is the density of the given material which has the usual unit of \[kg/{{m}^{3}}\].
If we substitute all these units to the expression of molar susceptibility,
\[\begin{align}
& {{\chi }_{m}}=\dfrac{M{{\chi }_{v}}}{\rho } \\
& \Rightarrow {{\chi }_{m}}=\dfrac{\dfrac{kg}{mol}}{\dfrac{kg}{{{m}^{3}}}}={{m}^{3}}/mol \\
\end{align}\]
So we got the unit of molar susceptibility as \[{{m}^{3}}/mol\] but it is actually represented as \[{{m}^{3}}\].
Therefore the correct answer is option c.
Note: Molar magnetic susceptibility is also used to classify between paramagnetic and diamagnetic materials. We must always keep in mind that volume susceptibility is a dimensionless quantity. Susceptibility can also be measured as mass susceptibility which has the unit of \[{{m}^{3}}/kg\].
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