What is the mass ratio of ethylene glycol (C_ | vedclass

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(C) Case $I$: For $0.25  m$ solution,molality $(m) = \frac{W_1 	imes 1000}{M_2 	imes W_{	ext{solvent}}}$.Assuming the mass of the solvent is approx

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(C) Case $I$: For $0.25  m$ solution,molality $(m) = \frac{W_1 	imes 1000}{M_2 	imes W_{	ext{solvent}}}$.Assuming the mass of the solvent is approximately equal to the mass of the solution $(500  g)$:$0.25 = \frac{W_1 	imes 1000}{62 	imes 500} \implies W_1 = \frac{0.25 	imes 62}{2} = 7.75  g$.Case $II$: For $0.25  M$ solution,molarity $(M) = \frac{W_2 	imes 1000}{M_2 	imes V_{	ext{solution}}}$.$0.25 = \frac{W_2 	imes 1000}{62 	imes 250} \implies W_2 = \frac{0.25 	imes 62}{4} = 3.875  g$.Ratio $\frac{W_1}{W_2} = \frac{7.75}{3.875} = \frac{2}{1}$.

What is the mass ratio of ethylene glycol $(C_2H_6O_2, \text{molar mass} = 62 g/mol)$ required for making $500 g$ of $0.25 m$ aqueous solution and $250 mL$ of $0.25 M$ aqueous solution?

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