The unit for molar susceptibility is

m^{3}

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k g - m^{- 3}

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k g^{- 1} - m^{3}

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Solution

The correct option is A $m^{3}$
Molar susceptibility
$= \frac{V o l u m e s u s c e p t i b i l i t y}{D e n s i t y o f m a t e r i a l} \times m o l e c u l a r w e i g h t$
$= \frac{\frac{I}{H}}{ρ} \times M = \frac{\frac{I}{H}}{\frac{M}{V}} \times M$
So it’s unit is $m^{3}$ .

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