# 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

Aug 6, 2017

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

#### Explanation:

We use a $1 \cdot m L$ volume of alcohol solution..........

We want $\text{Moles of ethyl alcohol"/"Volume of solution}$, we proceed on the basis of a $1 \cdot m L$ volume.........

$1 a .$ ((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

$1 b .$ We require $\text{solution molality"="moles of ethyl alcohol solute"/"kilograms of solvent}$

We already have the moles of solute in a $1 \cdot m L$ 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 $0.125 \cdot m o l$ of ethyl alcohol, but solution concentration is $2.14 \cdot m o l \cdot {L}^{-} 1$ with respect to ethyl alcohol, and so we takes the quotient.....

$\frac{0.125 \cdot m o l}{2.14 \cdot m o l \cdot {L}^{-} 1} = 0.0584 \cdot \frac{1}{{L}^{-} 1} = 0.0584 \cdot L \times {10}^{3} \cdot m L \cdot {L}^{-} 1 = 58.4 \cdot m L$

$d .$ 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 $1 \cdot m L$ volume..........

${\chi}_{{H}_{2} O} = \frac{\frac{0.984 \cdot g \times 0.9}{18.01 \cdot g \cdot m o {l}^{-} 1}}{\frac{0.984 \cdot g \times 0.9}{18.01 \cdot g \cdot m o {l}^{-} 1} + \frac{0.984 \cdot g \times 0.1}{46.07 \cdot g \cdot m o {l}^{-} 1}} = 0.958$

And since, by definition, ${\chi}_{{H}_{2} O} + {\chi}_{\text{ethanol}} = 1$

${\chi}_{\text{ethanol}} = 1 - {\chi}_{{H}_{2} O} = 1 - 0.958 = 0.042$.

You will have to check this calculation........

$\text{all care taken, but no responsibility admitted.........}$