We use a #1*mL# volume of alcohol solution..........
We want #"Moles of ethyl alcohol"/"Volume of solution"#, we proceed on the basis of a #1*mL# volume.........
#1a.# #((1*cancel(mL)xx0.984*cancel(g*mL^-1)xx10%)/(46.07*cancelg*mol^-1))/(1*cancel(mL)xx1xx10^-3*L*cancel(mL^-1))=2.14*mol*L^-1#
#1b.# We require #"solution molality"="moles of ethyl alcohol solute"/"kilograms of solvent"#
We already have the moles of solute in a #1*mL# volume, and the mass of solvent is something we can determine with #rho#.
#((1*cancel(mL)xx0.984*cancel(g*mL^-1)xx10%)/(46.07*cancelg*mol^-1))/(90%xx0.984*g*mL^-1xx1*mLxx10^-3*kg*g^-1)=2.42*mol*kg^-1#.
#c.# We want #0.125*mol# of ethyl alcohol, but solution concentration is #2.14*mol*L^-1# with respect to ethyl alcohol, and so we takes the quotient.....
#(0.125*mol)/(2.14*mol*L^-1)=0.0584*1/(L^-1)=0.0584*Lxx10^3*mL*L^-1=58.4*mL#
#d.# We want mole fraction with respect to water......
#chi_(H_2O)="moles of water"/("moles of water + moles of ethanol")#
We can work, again, from a #1*mL# volume..........
#chi_(H_2O)=((0.984*gxx0.9)/(18.01*g*mol^-1))/((0.984*gxx0.9)/(18.01*g*mol^-1)+(0.984*gxx0.1)/(46.07*g*mol^-1))=0.958#
And since, by definition, #chi_(H_2O)+chi_("ethanol")=1#
#chi_"ethanol"=1-chi_(H_2O)=1-0.958=0.042#.
You will have to check this calculation........
#"all care taken, but no responsibility admitted........."#
