If the density of $\mathrm{CH}_{3} \mathrm{OH}$ is $0.80 \mathrm{~kg} \mathrm{~L}^{-1}$, the volume of methanol to prepare $2.5 \mathrm{~L}$ of $0.25 \mathrm{M}$ aqueous solution is

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Aicharya · Accepted Answer

Since molarity, $M=n_{2} / V$, we have Amount of $\mathrm{CH}_{3} \mathrm{OH}$ required, $n_{2}=M V=\left(0.25 \mathrm{~mol} \mathrm{~L}^{-1}ight)(2.5 \mathrm{~L})=0.625 \mathrm{~mol}$ Mass of $\mathrm{CH}_{3} \mathrm{OH}$ required, $m_{2}=n_{2} M_{2}=(0.625 \mathrm{~mol})\left(32 \mathrm{~g} \mathrm{~mol}^{-1}ight)=20.0 \mathrm{~g}=20.0 \times 10^{-3} \mathrm{~kg}$ Volume of $\mathrm{CH}_{3} \mathrm{OH}$ required, $V=\frac{m_{2}}{ho}=\frac{20.0 \times 10^{-3} \mathrm{~kg}}{0.80 \mathrm{~kg} \mathrm{~L}^{-1}}=0.025 \mathrm{~L}=25.0 \mathrm{~mL}$ ^

If the density of \(\mathrm{CH}_{3} \mathrm{OH}\) is \(0.80 \mathrm{~kg} \mathrm{~L}^{-1}\), the volume of…

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