Number of g of oxygen in 32.2,g of Na_2SO_4 c | vedclass

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(B) The molar mass of $Na_2SO_4 \cdot 10H_2O$ is calculated as: $M = (2 	imes 23) + 32 + (4 	imes 16) + 10 	imes (2 	imes 1 + 16) = 46 + 32 + 64 +

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$22.4$

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(B) The molar mass of $Na_2SO_4 \cdot 10H_2O$ is calculated as: $M = (2 	imes 23) + 32 + (4 	imes 16) + 10 	imes (2 	imes 1 + 16) = 46 + 32 + 64 + 180 = 322\,g/mol$. In one mole $(322\,g)$ of $Na_2SO_4 \cdot 10H_2O$,the total mass of oxygen is: $O = (4 	imes 16) + (10 	imes 16) = 64 + 160 = 224\,g$. Therefore,in $32.2\,g$ of $Na_2SO_4 \cdot 10H_2O$,the mass of oxygen is: $	ext{Mass of oxygen} = \frac{32.2\,g}{322\,g} 	imes 224\,g = 0.1 	imes 224 = 22.4\,g$.

Number of $g$ of oxygen in $32.2\,g$ of $Na_2SO_4 \cdot 10H_2O$ is

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