Identical to the number of substances contained in water, 18 and 17 g H2S?

given:
m(Н2О) = 18 g
m(H2S) = 17 g
find:
ν(Н2О) - ?
ν(H2S) - ?
Solution:
1- way:
М(Н2О) = 18 g/mol
18 g Н2О  х mol Н2О
18 g Н2О 1 mol Н2О
х=(18 g  • 1 mol)/(18 g)   =1 mol
2- way:
М(Н2О) = 18 g/mol
ν=m/М  ;ν(Н₂О) =(m(Н₂О))/(М(Н₂О))  = (18 g)/(18 g/mol)  = 1 mol.
1- way:
М(H2S) = 34 g/mol
17g H2S  х mol H2S
34 g H2S 1 mol H2S
х= (17 g  • 1 mol)/(34 g)   =0,5 mol
2- way:
М(H2S) = 34 g/mol
ν=m/М  ;ν(H₂S) = m(H₂S)/М(H₂S)   = (17 g)/(34 g/mol)  = 0,5 mol.
Answer:ν(Н2О) = 1 mol; ν(H2S) =  0,5 mol - Misc.