If the density of methanol is 0.793 kg L^{- | vedclass

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(A) Density of methanol $= 0.793 \ kg \ L^{-1} = 793 \ g \ L^{-1}$.Molar mass of methanol $(CH_3OH) = 12.01 + 4(1.008) + 16.00 = 32.04 \ g \ mol^{-1}

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$25.22$

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(A) Density of methanol $= 0.793 \ kg \ L^{-1} = 793 \ g \ L^{-1}$.Molar mass of methanol $(CH_3OH) = 12.01 + 4(1.008) + 16.00 = 32.04 \ g \ mol^{-1} \approx 32 \ g \ mol^{-1}$.Molarity $(M_1)$ of the given methanol $= \frac{	ext{Density}}{	ext{Molar mass}} = \frac{793 \ g \ L^{-1}}{32 \ g \ mol^{-1}} = 24.78 \ M$.Using the dilution equation $M_1V_1 = M_2V_2$:$24.78 \ M 	imes V_1 = 0.25 \ M 	imes 2.5 \ L$.$V_1 = \frac{0.25 	imes 2.5}{24.78} \ L = 0.02522 \ L$.$V_1 = 0.02522 	imes 1000 \ mL = 25.22 \ mL$.

If the density of methanol is $0.793 \ kg \ L^{-1}$,what volume is needed to prepare $2.5 \ L$ of a $0.25 \ M$ solution (in $mL$)?

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