The concentration of ethanol in the solution called 86-proof vodka is \(6.5 \mathrm{M}\). If the density of…

The concentration of ethanol in the solution called 86-proof vodka is \(6.5 \mathrm{M}\). If the density of the solution is \(0.95 \mathrm{~g} \mathrm{~cm}^{-3}\), the amount fraction of ethanol in vodka is
  1. \(0.304\)
  2. \(0.252\)
  3. \(0.205\)
  4. \(0.152\)

Solution

For \(1 \mathrm{~L}\) of vodka, we have \(n_{2}=6.5 \mathrm{~mol}\) and \(V=1000 \mathrm{~cm}^{3}\) Mass of solution, \(m=V ho=\left(1000 \mathrm{~cm}^{3}ight)\left(0.95 \mathrm{~g} \mathrm{~cm}^{-3}ight)=950 \mathrm{~g}\)
Mass of ethanol, \(m_{2}=n_{2} M_{2}=(6.5 \mathrm{~mol})\left(46 \mathrm{~g} \mathrm{~mol}^{-1}ight)=299 \mathrm{~g}\)
Mass of water, \(m_{1}=m-m_{2}=950 \mathrm{~g}-299 \mathrm{~g}=651 \mathrm{~g}\)
Amount of water, \(n_{1}=\frac{m_{1}}{M_{1}}=\frac{651 \mathrm{~g}}{18 \mathrm{~g} \mathrm{~mol}^{-1}}=36.167 \mathrm{~mol}\)
Amount fraction of ethanol, \(x_{2}=\frac{n_{2}}{n_{1}+n_{2}}=\frac{6.5}{(36.167+6.5)}=0.152\) undefined

Asked in: JEE-TOPICTESTS-CHEMISTRY

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