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ICSE Class 10 Chemistry 2020 Paper

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Calculate the percentage of :
in sodium aluminium fluoride [Na3AlF6][\mathrm{Na}_3\mathrm{AlF}_6], to the nearest whole number.
[Atomic Mass : Na=23, Al=27, F=19\mathrm{Na}=23,\ \mathrm{Al}=27,\ \mathrm{F}=19 ]
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Question : 66 of 86
Marks: +1, -0
Fluorine
Solution:  
Molecular Mass of Na3AlF6\mathrm{Na}_3\mathrm{AlF}_6
=23×3+27+19×6=210=23 \times 3+27+19 \times 6=210
Percentage of fluorine
=  6×19210×100=54.28≃54%=\;\frac{6 \times 19}{210} \times 100=54.28 \simeq 54\%
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