3.42 g of sucrose are dissolved in 18g of water in a beaker. The number of oxygen atoms in the solution are (a) 6.68 × 1023 (b) 6.09 × 1022 (c) 6.022 × 1023 (d) 6.022 × 1021

Answer: (a) 6.68 × 1023

  • Carbon = 12
  • Hydrogen = 1
  • Oxygen = 16

C = 12 × 12 = 144
H = 1 × 22 = 22
O = 16 × 11 = 176

Sucrose = 144 + 22 + 176 = 342 g/mol

1 mol of sucrose (C12H22O11) contains = 11× NA atoms of oxygen,

where NA = 6.023×1023

0.01 mol of sucrose (C12H22O11) contains = 0.01 × 11 × NA atoms of oxygen

= 0.11× NA atoms of oxygen

= 18 g/(1 × 2+ 16)gmol-1

=18 g /18 gmol-1

= 1mol

1mol of water (H2O) contains 1 × NA atom of oxygen

Total number of oxygen atoms = Number of oxygen atoms from sucrose + Number of oxygen atoms from water

Total number of oxygen atoms = 0.11 NA + 1.0 NA = 1.11NA

Number of oxygen atoms in solution = 1.11 × Avogadro’s number

Number of oxygen atoms in solution = 1.11 × 6.022 ×1023

Number of oxygen atoms in solution = 6.68 × 1023

Was this answer helpful?

 
   

0 (0)

(0)
(0)

Choose An Option That Best Describes Your Problem

Thank you. Your Feedback will Help us Serve you better.

Leave a Comment

Your Mobile number and Email id will not be published. Required fields are marked *

*

*

BOOK

Free Class

Ask
Question