Calculate the number of moles of helium present in $6.46 g$ . (The atomic weight of helium is $4$ a.m.u.).

16.15

moles

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1.615

moles

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161.5

moles

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0.1615

moles

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Solution

The correct option is B $1.615$ moles
given that
$6.46 g$ Helium
moles $= \frac{W e i g h t o f H e}{m o l e c u l a r w e i g h t o f H e}$
$= \frac{6.46}{4}$
$= 1.615 m o l e s$

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