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(D) The molar mass of $Na_2SO_4 \cdot 10H_2O$ is calculated as: $2(23) + 32 + 4(16) + 10(18) = 46 + 32 + 64 + 180 = 322 \ g/mol$. One mole of $Na_2SO_

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$2.8$

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(D) The molar mass of $Na_2SO_4 \cdot 10H_2O$ is calculated as: $2(23) + 32 + 4(16) + 10(18) = 46 + 32 + 64 + 180 = 322 \ g/mol$. One mole of $Na_2SO_4 \cdot 10H_2O$ contains $4$ oxygen atoms from $SO_4^{2-}$ and $10$ oxygen atoms from $10H_2O$,totaling $14$ moles of oxygen atoms. The number of moles of $Na_2SO_4 \cdot 10H_2O$ in $64.4 \ g$ is: $n = \frac{64.4 \ g}{322 \ g/mol} = 0.2 \ mol$. Therefore,the moles of oxygen atoms = $0.2 \ mol 	imes 14 = 2.8 \ mol$.

How many moles of oxygen atoms are present in $64.4 \ g$ of $Na_2SO_4 \cdot 10H_2O$?

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