question_answer 34)

  3.42 g of sucrose are dissolved in 18 g of water in a beaker. The number of oxygen atoms in the solution are (a) 6.68 x1023                      (b) 6.09 x1022                     (c) 6.022 x 10                      (d) 6.022 x1021

Answer:

  (a)  Step 1. Molar mass of sucrose,   or                   342 g = 1 mole of sucrose                        3,42 g = 0.01 mole of sucrose 1 mole of sucrose  contains 0-atoms = 11 x 6.022 x1023 atoms  0.01 mole of sucrose will contain 0-atoms = 0.01 x 11 x 6.022x 1023 atoms = 6.6242 x 1022 Step 2. 18 g of water  = 1 mole of water 1 mole of water  contains 0-atoms = 6.022 x 1023 atoms Step 3. By adding the number of I-atoms present in 3.42 g of sucrose and 18 g of water we get 6.022x 1023 + 6.6242 x 1022 = 1022 (60,22 +6.6242) = 66.844 x 1022 = 6.68x 1023 atoms