The density of an aqueous solution containing 10% of ethanol, C2H5OH, by mass is 0.984 g/mL. Calculate the a) M b) m of this equation c) What volume of the solution would contain 0.125 mole of ethanol? d) Calculate the mole fraction of water inthesolution

1 Answer
Aug 6, 2017

Answer:

Is that all you want.......?

Explanation:

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........."#