The mass of ethanol (molar mass = $46 \mathrm{~g} \mathrm{~mol}^{-1}$ ) to be added to $1.0 \mathrm{~kg}$ of…
The mass of ethanol (molar mass = $46 \mathrm{~g} \mathrm{~mol}^{-1}$ ) to be added to $1.0 \mathrm{~kg}$ of water so as to have its amount fraction equal to $0.2$ is
$319.5 \mathrm{~g}$
$432.1 \mathrm{~g}$
$638.9 \mathrm{~g}$
$719.3 \mathrm{~g}$
Solution
\(\begin{aligned} & x=\frac{\text { No. of moles of Ethanol }}{\text { No. of moles of ethanol }+ \text { No. of moles of water }} \\ & 0.2=\frac{\left(\frac{\mathrm{m}}{46} \mathrm{~g} \mathrm{~mol}^{-1}ight)}{\left(\frac{\mathrm{m}}{46} \mathrm{~g} \mathrm{~mol}^{-1}ight)+\left(1000 \mathrm{~g} / 18 \mathrm{~g} \mathrm{~mol}^{-1}ight)} \\ & \Rightarrow 0.2 \times\left(\left(\frac{\mathrm{m}}{46} \mathrm{~g} \mathrm{~mol}^{-1}ight)+\left(1000 \frac{\mathrm{g}}{18} \mathrm{~g} \mathrm{~mol}^{-1}ight)ight)=\left(\frac{\mathrm{m}}{46} \mathrm{~g} \mathrm{~mol}^{-1}ight) \\ & \Rightarrow \frac{\mathrm{m}}{46}=\frac{0.2 \times 1000}{18} \times \frac{1}{0.8} \\ & \frac{\mathrm{m}}{46}=13.89 \\ & \Rightarrow \mathrm{m}=13.89 \times 46 \mathrm{~g}=638.89 \mathrm{~g}\end{aligned}\)
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