Given that 32.0g sulphur contains 6.02 x 10 23 sulphur atoms how many atoms are there in 2.70g of aluminium? {Al = 27, S =32}

Omejeb

8 years ago

1g of sulphur is equals to Avogadro's constant;6.02 then 0.1g of Al is equals to ----.then if more less divide..0.1x 6.02x1021then the answer becomes 6.02x1022

achason

7 years ago

n=m/M i.e no of moles equals given mass(m) over molar mass(M)
n=2.7/27
n=0.1
then 6.02x10^23 x 0.1=6.02x10^22 atoms

Gabbiola

7 years ago

The correct answer is C

Heritage07

4 years ago

The correct answer is D
Since sulphur, 32 is given to be 6.02×10^23, and aluminium is 2.70(the unknown), cross multiplying
So 2.7 × Avogadro's constant, 6.02 ×10^23 divided by 32. Equals 5.08 ×10^22.
Unlike what is done, where aluminium is 6.02×10^23, that's incorrect.