Correct Answer - `6.022 xx 10^(22)` molecules
Gram molecular mass of can sugar `(C_(12)H_(22)O_(11))=12xx12+22xx1+16xx11=342.0g`
342.0 g of oxygen contain `= 6.022 xx 10^(23)` molecules
34.2 g of oxygen contain `= (34.2)/(342)xx6.022xx10^(23)=6.022xx10^(22)` molecules
