The molality of the solution prepared by diss | vedclass

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(C) Step $1$: Calculate the mass of methanol (solute).Mass = Density $	imes$ Volume = $0.8 \ g \ mL^{-1} 	imes 125 \ mL = 100 \ g$.Step $2$: Calcula

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$8.33$

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(C) Step $1$: Calculate the mass of methanol (solute).Mass = Density $	imes$ Volume = $0.8 \ g \ mL^{-1} 	imes 125 \ mL = 100 \ g$.Step $2$: Calculate the moles of methanol $(CH_3OH)$.Molar mass of $CH_3OH = 12 + 4(1) + 16 = 32 \ g \ mol^{-1}$.Moles of $CH_3OH = \frac{100 \ g}{32 \ g \ mol^{-1}} = 3.125 \ mol$.Step $3$: Calculate the molality $(m)$.Molality = $\frac{	ext{moles of solute}}{	ext{mass of solvent in kg}}$.Mass of solvent (ethanol) = $375 \ g = 0.375 \ kg$.$m = \frac{3.125 \ mol}{0.375 \ kg} \approx 8.33 \ mol \ kg^{-1}$.

The molality of the solution prepared by dissolving $125 \ mL$ of pure methanol $(d = 0.8 \ g \ mL^{-1})$ in $375 \ g$ of ethanol is

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