Match the Van der Waals constant a (in L^2 cd | vedclass

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(C) The Van der Waals constant $a$ represents the magnitude of intermolecular attractive forces.Larger and more complex molecules have stronger interm

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$1-c, 2-d, 3-a, 4-b$

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(C) The Van der Waals constant $a$ represents the magnitude of intermolecular attractive forces.Larger and more complex molecules have stronger intermolecular forces,leading to higher $a$ values.Comparing the given gases:$1. C_6H_6$ (Benzene) has $a \approx 18.000$.$2. C_6H_5CH_3$ (Toluene) is larger than benzene,so it has the highest $a \approx 24.060$.$3. Ne$ (Neon) is a noble gas with weak dispersion forces,so it has the lowest $a \approx 0.217$.$4. H_2O$ (Water) has hydrogen bonding,giving $a \approx 5.464$.Thus,the correct matching is: $1-c, 2-d, 3-a, 4-b$.

Gas	$a$ value
$1. C_6H_{6(g)}$	$a. 0.217$
$2. C_6H_5CH_{3(g)}$	$b. 5.464$
$3. Ne_{(g)}$	$c. 18.000$
$4. H_2O_{(g)}$	$d. 24.060$

Match the Van der Waals constant $a$ (in $L^2 \cdot bar \cdot mol^{-2}$) for the following gases:

Gas $a$ value

$1. C_6H_{6(g)}$ $a. 0.217$

$2. C_6H_5CH_{3(g)}$ $b. 5.464$

$3. Ne_{(g)}$ $c. 18.000$

$4. H_2O_{(g)}$ $d. 24.060$

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Match the Van der Waals constant $a$ (in $L^2 \cdot bar \cdot mol^{-2}$) for the following gases: Gas $a$ value $1. C_6H_{6(g)}$ $a. 0.217$ $2. C_6H_5CH_{3(g)}$ $b. 5.464$ $3. Ne_{(g)}$ $c. 18.000$ $4. H_2O_{(g)}$ $d. 24.060$