Question

5. How many carbon atoms are present in 34.2g of sucrose (C₁2H22O11) M.MASS = 342g/mol)? A) 6.0 × 1022 B) 7.2 x 1023 C) 3.6 × 1025 D) 3.6 × 1024​

Answer 1

MwMw of sucrose =C12H22O11=C12H22O11

=12×12+1×22+1611=12×12+1×22+1611

=342g=342g

a. 1 mol of sucrose =342g=12molofC=342g=12molofC

∴3.42g=12342×3.42∴3.42g=12342×3.42

=0.12molofC=0.12×6.023×1023atoms=0.12molofC=0.12×6.023×1023atoms

=7.228×1022atoms ofC=7.228×1022atoms ofC

b. 342g342g of Sucrose =22molofH=22molofH

3.42g3.42g of sucrose =22342×3.42=22342×3.42

=0.22molofH=0.22molofH

=0.22×6.023×1023=0.22×6.023×1023atoms

=132.5×1021=132.5×1021 atom of HH

c. 342g342g of sucrose =11×6.023×1023342×3.42=11×6.023×1023342×3.42

=66.25×1021=66.25×1021 atoms of O