3 years ago

How many fluorine atoms are present in 125.0g of phosphorus pentafluoride?

1 answer:

8 0

molar mass of PF5 = 125.966 g/mol

125 g PF5 × (1 mol PF5/125.966 g PF5) = 0.992 mol PF5

0.992 mol PF5 × (6.022 × 10^23 molecules PF5);

= 5.97 × 10^23 molecules PF5

Since there 5 fluorine atoms per molecule of PF5,

(5.97.× 10^23 molecules PF5) × (5 atoms F/1 molecule PF5)

= 2.99 × 10^24 atoms F

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