# When 50 mL of liquid water at 25 °C is added to 50 mL of ethanol (ethyl alcohol), also at 25 °C, the combined volume of the mixture is considerably less than 100 mL. What is a possible explanation for this?

Mar 28, 2018

Alcohol and water are no-additive volumes.

#### Explanation:

Alcohol and water are no-additive volumes due that alcohol is soluble in water; alcohol molecules occupies water voids via H-bonds.

Here are two ways to calculate present question ( I am not going to explain the theory):

$\textcolor{g r e e n}{\text{FROM COMPONENTS AND MIXTURE DENSITIES}}$ @ $25 C$
(Water density = 0.99705 g/cc; Ethanol density = 0.78522 g/cc)

Data,
${V}_{\text{total}} = 100 m L$

Water grams = .99705 g/"ml" (50 "ml")=49.8525

Ethanol grams = .78522 g/"ml" (50 "ml")$\left(\frac{1}{46} \frac{\text{mol}}{g}\right)$ = 39.261

Water %wt = 49.8525/(49.8525+ 39.261)= 55.9427

Ethanol %wt = 100 - 55.9427 = 44.0573

From literature can be found the MIXTURE DENSITY @ 44%wt (25 C) of Ethanol $\implies$ 0.92571 g/ml

Finally, for two mixed volumes of 50 ml/each:

$\textcolor{b l u e}{\text{Volume of mixture}}$ = "(55.9427+44.0573)g"/( 0.92571 g/"ml")$=$$\textcolor{b l u e}{\text{96.2650 ml}}$

$\textcolor{b r o w n}{\text{FROM EACH COMPONENT MOLAR VOLUME}}$
(Water = 17.9 ml/mol; Ethanol = 55 ml/mol)

Water moles = .99705 g/"ml" (50 "ml")$\left(\frac{1}{18} \frac{\text{mol}}{g}\right)$ = 2.769583

Ethanol moles = .78522 g/"ml" (50 "ml")#$\left(\frac{1}{46} \frac{\text{mol}}{g}\right)$ = 0.8535

$\textcolor{b l u e}{\text{Volume of mixture}}$ = $\left(17.9 \text{ml"/"mol") 2.769583 mol +(55 "ml"/"mol}\right) 0.8535 m o l$$=$$\textcolor{b l u e}{\text{96.5180 ml}}$