Question

What is the weight of 6.023*10^22 sulphur atom?​

Answer 1

Answer:

1 gram atom = N – atoms = 6.023 x 1023 atoms = gram atomic weight

gram atomic weight = weight of  N atoms in grams

1 gram molecule = N- molecules = 6.023 x 1023 molecules = gram molecular wt.

gram molecular wt. = wt. of N molecules in gram

number of moles of molecules ‘n’ = wt. (gram)/ molecular wt.= w/ m

number of moles of atoms ‘n’ = wt. (gram)/ atomic wt.

no. of moles ‘n’ = no. of particles / ( 6.023 x 1023)

Answer 2

Answer:

How many gram atoms of Sulphur and how many gram atoms of oxygen are needed to prepare 6.023×10^24 molecules of SO2?

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Gram atoms of sulphur

and gram atoms of oxygen ????

1 mol SO2 contains 6.0213*10^23 molecules

6.023*10^24 molecules = 10 mol SO2

Equation

S(s) + O2(g) → SO2(g)

1 mol S reacts with 1 mol O2 to prepare 1 mol SO2

To prepare 10 mol SO2 you require : 10 mol S plus 10 mol O2

Molar mass S = 32 g/mol You require 10 mol = 320 g

Molar mass O2 = 32 g/mol :You require 10 mol = 320 g