Question

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?

Answer 1

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):

`color(green)"FROM COMPONENTS AND MIXTURE DENSITIES"` @ `25 C`
(Water density = 0.99705 g/cc; Ethanol density = 0.78522 g/cc)

Data,
`V_"total" = 100 mL`

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

Ethanol grams = .78522 `g/"ml" (50 "ml")` `(1/46 "mol"/g)` = 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 `=>` 0.92571 g/ml

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

`color(blue)"Volume of mixture"` = `"(55.9427+44.0573)g"/( 0.92571 g/"ml")` `=` `color(blue)"96.2650 ml"`

`color(brown)"FROM EACH COMPONENT MOLAR VOLUME"`
(Water = 17.9 ml/mol; Ethanol = 55 ml/mol)

Water moles = .99705 `g/"ml" (50 "ml")` `(1/18 "mol"/g)` = 2.769583

Ethanol moles = .78522 `g/"ml" (50 "ml")` `(1/46 "mol"/g)` = 0.8535

`color(blue)"Volume of mixture"` = `(17.9 "ml"/"mol") 2.769583 mol +(55 "ml"/"mol") 0.8535 mol` `=` `color(blue)"96.5180 ml"`