(a) Gram molecular mass of `H_(2)SO_(4)=2xx1+32+4xx16=98g`
`98g of H_(2)SO_(4)=1"gram molecules"`
`4.9"g of "H_(2)SO_(4)=(4.9)/(98)=0.05"gram molecule"`
(b) `underset("1mol")(H_(2)O) =underset("2 gram aroms")(2H)+underset(1"grams atom")(O)`
1 mole of water `(H_(2)O)` has H atoms=2grams
0.15 moles of water `(H_(2)O)` has H atoms=`(2xx0.15)=0.30grams`
`=0.30xx6.022xx10^(23)"atoms"`
`=1.81xx10^(23)"atoms"`
1 mole of water `(H_(2)O)` has O atoms =1 gram atoms
0.015 moles of water `(H_(2)O)` has O atoms=0.15gram atoms
`=0.15xx6.022xx10^(23)=9.03xx10^(23)"atoms"`