Chemistry

(a) Give a difference between ionization and electrolytic dissociation.
(b) 112 mL of a gaseous fluoride of a non-metal Phosphorus at S.T.P. has a mass of 0.63 g. Calculate the relative molecular mass of the fluoride.
(c) If this compound given in (b) has only one atom of Phosphorus, then determine its formula. (At. Wt. P=31, F=19)

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(a)

Ionisation	Electrolytic dissociation
Formation of positively or negatively charged ions from molecules which are not initially in the ionic state.	Separation of ions which are already present in an ionic compound

(b) Given,

112 mL of gaseous fluoride has mass = 0.63 g

∴ 22400 cm³ of gaseous fluoride will have mass = $\dfrac{0.63}{112}$ x 22400

= 126 g

∴ Relative molecular mass of fluoride = 126 g

Atomic mass of P + n × atomic mass of F = 126

⇒ 31 + 19n = 126

⇒ 19n = 126 - 31

⇒ 19n = 95

⇒ n = $\dfrac{95}{19}$

⇒ n = 5

∴ The molecular formula is PF₅

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