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Calculate the excess molar volume $(V_E)$ of a binary mixture of benzene and toluene at 298 K, given that the partial molar volumes of benzene and toluene at infinite dilution are $89.7 cm^3/mol$ and $103.0 cm^3/mol$ , respectively, and the molar volumes of the pure components at 298 K are $126.6 cm^3/mol$ and $147.1 cm^3/mol$ , respectively. Assume that the mixture behaves as ideal binary mixture.
Option: 1
$-3.4 cm^3/mol$

Option: 2
$-2.6 cm^3/mol$

Option: 3
$2.6 cm^3/mol$

Option: 4
$3.4 cm^3/mol$

Answers (1)

best_answer

The excess molar volume $(V_E)$ of an ideal binary mixture is given by the equation:

$V_E = V - (x_AV_A + x_BV_B)$

where V is the molar volume of the mixture, $V_A$ and $V_B$ are the molar volumes of the pure components, and $x_A$ and $x_B$ are the mole fractions of the components A and B in the mixture.

Using the given data, we have:

$x_A = 1/2$ (since it is a binary mixture)

$x_B = 1/2$

$V_A = 126.6 cm^3/mol$
$V_B = 147.1 cm^3/mol$
$V = x_AV_A + x_BV_B = 136.9 cm^3/mol$

At infinite dilution, the partial molar volume of A and B are:

$V_A\ ^\infty = 89.7 cm^3/mol$
$V_B\ ^\infty = 103.0 cm^3/mol$

The excess molar volume can now be calculated as:
$V_E = V - (x_AV_A\ ^\infty + x_BV_B\ ^\infty)\\\\ = 136.9 cm^3/mol - [1/2(89.7 cm^3/mol) + 1/2(103.0 cm^3/mol)]\\\\ = -3.4 cm^3/mol$

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seema garhwal

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