Consider a solution of ethanol in water in which the mole fraction of ethnol is `0.040`. If the density of `H_(2)O` is taken to be `1 g//mL` the molarity of ethanol in solution is

Options

A. `3.41 M`
B. `2.31 M`
C. `4.75 M`
D. `1.78 M`

Answer

2

Explanation

Molarity is equal to number of moles of `C_(2)H_(5)OH` in `1 L` of the solution. Since the density of solution is not provided and the density of `H_(2)O` (which has dissolved `C_(2)H_(5)OH)` is taken to be `1 g//mL` message is that the solution is dilute to such an extent that when `C_(2)H_(5)OH` is dissolved in `H_(2)O`, its density `(1 g mL^(-1))` remains practically unchanged.
`chi_(C_(2)H_(5)OH)=n_(C_(2)H_(5)OH)/n_("total")=0.040=4/100`
Considering 100 moles of soln, we have
`n_(C_(2)H_(5)OH)=4`
`n_(H_(2)O)=96`
Since solution is dilute,
`(Volume)_(soln)=(Volume)_(H_(2)O)=(mass_(H_(2)O))/d_(H_(2)O)`
`=((n_(H_(2)O))(molar mass_(H_(2)O)))/d_(H_(2)O)`
`=((96 mol)(18 g mol^(-1)))/(1 g mL^(-1))`
`=1728 mL`
Now, molarity `(M)=n_(C_(2)H_(5)OH)/V_(mL)xx(1000 mL)/L`
`=(4mol C_(2)H_(5)OH)/(1728 mL)xx(1000 mL)/L`
`=2.31 mol L^(-1)`