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?

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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?

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Answer 1 · 2018-03-28T04:14:43

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):

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

Data,
$V_{total} = 100 m L$ $V_"total" = 100 mL$

Water grams = .99705 $\frac{g}{ml} (50 ml)$ $g/"ml" (50 "ml")$ =49.8525

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

$W a t e r % w t = \frac{49.8525}{49.8525 + 39.261}$ $Water %wt = 49.8525/(49.8525+ 39.261)$ #= 55.9427

$E t h a n o l % w t = 100 - 55.9427$ $Ethanol %wt = 100 - 55.9427$ = 44.0573

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

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

$Volume of mixture$ $color(blue)"Volume of mixture"$ = $\frac{(55.9427+44.0573)g}{0.92571 \frac{g}{ml}}$ $"(55.9427+44.0573)g"/( 0.92571 g/"ml")$ $=$ $=$ $96.2650 ml$ $color(blue)"96.2650 ml"$

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

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

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

$Volume of mixture$ $color(blue)"Volume of mixture"$ = $(17.9 \frac{ml}{mol}) 2.769583 m o l + (55 \frac{ml}{mol}) 0.8535 m o l$ $(17.9 "ml"/"mol") 2.769583 mol +(55 "ml"/"mol") 0.8535 mol$ $=$ $=$ $96.5180 ml$ $color(blue)"96.5180 ml"$

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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?

1 Answer

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