3.0, g of oxalic acid [(CO_2H)_2 cdot 2H_2O] | vedclass

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(C) Step $1$: Calculate the mass of the solution. $	ext{Mass of solution} = 	ext{Volume} 	imes 	ext{Density} = 250\, mL 	imes 1.9\, g/mL = 475\,

Vedclass · Accepted Answer

$0.05\, mol\, kg^{-1}$ and $0.19\, N$

Vedclass · Answer

(C) Step $1$: Calculate the mass of the solution. $	ext{Mass of solution} = 	ext{Volume} 	imes 	ext{Density} = 250\, mL 	imes 1.9\, g/mL = 475\, g$. Step $2$: Calculate the mass of the solvent. $	ext{Mass of solvent} = 	ext{Mass of solution} - 	ext{Mass of solute} = 475\, g - 3.0\, g = 472\, g$. Step $3$: Calculate molality. $	ext{Molar mass of } (CO_2H)_2 \cdot 2H_2O = 126\, g\, mol^{-1}$. $	ext{Molality} = \frac{3.0 	imes 1000}{126 	imes 472} \approx 0.05\, mol\, kg^{-1}$. Step $4$: Calculate normality. $	ext{Equivalent mass of oxalic acid} = \frac{126}{2} = 63\, g/	ext{equiv}$. $	ext{Normality} = \frac{	ext{Mass of solute} 	imes 1000}{	ext{Equivalent mass} 	imes 	ext{Volume of solution } (mL)} = \frac{3.0 	imes 1000}{63 	imes 250} = 0.19\, N$.

$3.0\, g$ of oxalic acid $[(CO_2H)_2 \cdot 2H_2O]$ is dissolved in a solvent to prepare a $250\, mL$ solution. The density of the solution is $1.9\, g/mL$. The molality and normality of the solution,respectively,are closest to

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