One mole of PCl5 (208.5 g/mol) is 25% dissociated into PCl3 and Cl2. Choose the correct statement(s).

The total number of moles at equilibrium is 1.25.

The average molar mass of the mixture is 166.8 g/mol.

If dissociation were complete, the vapour density would be 52.125.

The total number of moles at equilibrium is 1.25.

The average molar mass of the mixture is 166.8 g/mol.

The vapour density of the mixture is 166.8.

If dissociation were complete, the vapour density would be 52.125.

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JEE Advanced levelOne or more correct optionsDifficulty 3 of 3mc107

One mole of $PCl X_{5}$ ( $208.5 g/mol$ ) is $25%$ dissociated into $PCl X_{3}$ and $Cl X_{2}$ . Choose the correct statement(s).

AThe total number of moles at equilibrium is $1.25$ .
BThe average molar mass of the mixture is $166.8 g/mol$ .
CThe vapour density of the mixture is $166.8$ .
DIf dissociation were complete, the vapour density would be $52.125$ .

Answer: A, B, D

Moles = 1 + 0.25 = 1.25 (A true). Average molar mass = 208.5/1.25 = 166.8 g/mol (B true); vapour density is half of that, 83.4, so C is false. For complete dissociation moles = 2, M = 104.25, vapour density = 52.125 (D true).

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One mole of PClX5​ (208.5 g/mol) is 25% dissociated into PClX3​ and ClX2​. Choose the correct statement(s).

Answer: A, B, D

More questions on Average molar mass of mixtures, vapour density

One mole of $PCl X_{5}$ ( $208.5 g/mol$ ) is $25%$ dissociated into $PCl X_{3}$ and $Cl X_{2}$ . Choose the correct statement(s).